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SeanforfunSeanforfun
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[Function add]
1. Add leetcode solutions.
1 parent 8f0f41b commit c475a62

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lines changed

Diff for: ‎leetcode/22. Generate Parentheses.md

+22-1
Original file line numberDiff line numberDiff line change
@@ -106,4 +106,25 @@ class Solution {
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}
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}
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}
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```
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```
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### Fourth time
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* Method 1: recursion
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```Java
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class Solution {
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private List<String> result;
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public List<String> generateParenthesis(int n) {
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this.result = new ArrayList<>();
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if(n <= 0) return result;
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dfs("", n, n);
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return result;
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}
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private void dfs(String cur, int left, int right){
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if(left == 0 && right == 0) result.add(new String(cur));
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else{
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if(left > 0) dfs(cur + "(", left - 1, right);
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if(right > left) dfs(cur + ")", left, right - 1);
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}
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}
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}
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```

Diff for: ‎leetcode/268. Missing Number.md

+23-2
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@@ -39,7 +39,7 @@ class Solution {
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}
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```
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###Second time
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### Second time
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1. We first get the sum of the array and the missing value is the deduction between expect sum and sum.
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```Java
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class Solution {
@@ -52,4 +52,25 @@ class Solution {
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return expect - sum;
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}
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}
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```
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```
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### Third Time
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* Method 1: Another method to find missing number in a sorted array.
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```Java
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class Solution {
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public int missingNumber(int[] nums) {
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Arrays.sort(nums);
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int low = 0, high = nums.length - 1;
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return binarySearch(nums, low, high);
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}
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private int binarySearch(int[] nums, int low, int high){
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int mid = low + (high - low) / 2;
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if(nums[mid] == mid && nums[mid + 1] != mid + 1) return mid + 1;
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else if(nums[mid] != mid){
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return binarySearch(nums, low, mid - 1);
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}else{
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return binarySearch(nums, mid + 1, high);
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}
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}
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}
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```

Diff for: ‎leetcode/273. Integer to English Words.md

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Original file line numberDiff line numberDiff line change
@@ -117,4 +117,59 @@ Output: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Th
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return result;
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}
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}
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```
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### Third Time
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* Method 1: String
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```Java
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class Solution {
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private static final String[] units = new String[]{"", "Thousand", "Million", "Billion"};
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private static final String[] ones = new String[]{"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"};
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private static String[] tens = new String[]{"", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
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private static String[] overTens = new String[]{"", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
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public String numberToWords(int num) {
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if(num == 0) return "Zero";
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String res = "";
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int bigUnitIndex = 0;
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while(num > 0){
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int cur = num % 1000;
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String curString = parse(cur);
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num /= 1000;
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if(bigUnitIndex == 0){
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res = curString;
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++bigUnitIndex;
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continue;
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}
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if(curString.length() != 0){
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res = curString + " " + units[bigUnitIndex] + (res.length() == 0 ? "": " ") + res;
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}
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bigUnitIndex++;
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}
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return res;
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}
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private String parse(int num){
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if(num < 10) return ones[num];
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else if(num > 10 && num < 20) return overTens[num - 10];
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else{
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String res = "";
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if(num % 100 > 10 && num % 100 < 20){
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res = overTens[num % 100 - 10];
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num /= 100;
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}else{
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if(num % 10 != 0){
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res = ones[num % 10];
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}
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num /= 10;
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if(num % 10 != 0){
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res = tens[num % 10] + (res.length() == 0 ? "": " " + res);
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}
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num /= 10;
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}
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if(num % 10 != 0){
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res = ones[num % 10] + " Hundred" + (res.length() == 0 ? "": " ") + res;
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}
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return res;
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}
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}
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}
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```

Diff for: ‎leetcode/344. Reverse String.md

+19-1
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@@ -55,4 +55,22 @@ class Solution {
5555
arr[j] = temp;
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}
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}
58-
```
58+
```
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### Second Time
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* Method 1: in-place, 2 pointers
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```Java
63+
class Solution {
64+
public void reverseString(char[] s) {
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int low = 0, high = s.length - 1;
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while(low < high){
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swap(s, low++, high--);
68+
}
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}
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private void swap(char[] s, int a, int b){
71+
char temp = s[a];
72+
s[a] = s[b];
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s[b] = temp;
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}
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}
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```

Diff for: ‎leetcode/93. Restore IP Addresses.md

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@@ -40,3 +40,31 @@ Output: ["255.255.11.135", "255.255.111.35"]
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}
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}
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```
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### Amazon Session
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* Method 1: recursion
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```Java
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class Solution {
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private List<String> result;
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private String s;
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public List<String> restoreIpAddresses(String s) {
51+
this.result = new ArrayList<>();
52+
this.s = s;
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dfs(0, 0, "");
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return this.result;
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}
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private void dfs(int index, int count, String temp){
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if(index == s.length() && count == 4) this.result.add(temp);
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else if(count < 4){
59+
for(int i = index + 1; i <= index + 3 && i <= s.length(); i++){
60+
String sub = s.substring(index, i);
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int cur = Integer.parseInt(sub);
62+
if(sub.charAt(0) == '0' && sub.length() != 1) return;
63+
if(cur >= 0 && cur <= 255){
64+
dfs(i, count + 1, temp + (temp.length() == 0 ? "": '.') + cur);
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}
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}
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}
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}
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}
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```

Diff for: ‎leetcode/94. Binary Tree Inorder Traversal.md

+29-1
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@@ -28,4 +28,32 @@ class Solution {
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inorder(n.right, result);
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}
3030
}
31-
```
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```
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### Second time
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* Method 1: Recursion
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```Java
36+
/**
37+
* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
43+
* }
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*/
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class Solution {
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private List<Integer> result;
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public List<Integer> inorderTraversal(TreeNode root) {
48+
this.result = new ArrayList<>();
49+
dfs(root);
50+
return this.result;
51+
}
52+
private void dfs(TreeNode node){
53+
if(node == null) return;
54+
dfs(node.left);
55+
this.result.add(node.val);
56+
dfs(node.right);
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}
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}
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```

Diff for: ‎leetcode/97. Interleaving String.md

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@@ -79,3 +79,38 @@ class Solution {
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}
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}
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```
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### Third time
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* Method 1: dp
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```Java
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class Solution {
87+
public boolean isInterleave(String s1, String s2, String s3) {
88+
char[] arr1 = s1.toCharArray();
89+
char[] arr2 = s2.toCharArray();
90+
char[] arr3 = s3.toCharArray();
91+
if(arr1.length + arr2.length != arr3.length) return false;
92+
boolean[][] dp = new boolean[arr1.length + 1][arr2.length + 1];
93+
dp[0][0] = true;
94+
for(int i = 0; i < arr1.length; i++){
95+
if(arr1[i] == arr3[i])
96+
dp[i + 1][0] = dp[i][0];
97+
}
98+
for(int i = 0; i < arr2.length; i++){
99+
if(arr2[i] == arr3[i])
100+
dp[0][i + 1] = dp[0][i];
101+
}
102+
for(int i = 1; i <= arr1.length; i++){
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for(int j = 1; j <= arr2.length; j++){
104+
// dp[i][j] saves if i + j - 1 can be constructed by i && j
105+
if(arr3[i + j - 1] == arr1[i - 1]){
106+
dp[i][j] = dp[i - 1][j];
107+
}
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if(arr3[i + j - 1] == arr2[j - 1]){
109+
dp[i][j] |= dp[i][j - 1];
110+
}
111+
}
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}
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return dp[arr1.length][arr2.length];
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}
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}
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```

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