279. Perfect Squares.md

279. Perfect Squares

Question

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.

Example 1:

Input: n = 12
Output: 3
Explanation: 12 = 4 + 4 + 4.

Example 2:

Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.

Thinking:

• Method 1: dp

class Solution {
    public int numSquares(int n) {
        int[] dp = new int[n + 1];
        for(int i = 1; i <= n; i++){
            int res = Integer.MAX_VALUE;
            for(int j = 1; j * j <= i; j++){
                res = Math.min(res, dp[i - j * j] + 1);
            }
            dp[i] = res;
        }
        return dp[n];
    }
}

Second time

• Method 1 dp

  1. Set all integer squares smaller than n to 1.
  2. Initial all other values to MAX_VALUE.
  3. for each number, if dp[i] == 1, skip, else check all add combination of the values and take the minimum value.

  class Solution {
      public int numSquares(int n) {
          int[] dp = new int[n + 1];
          for(int i = 0; i <= n; i++){
              dp[i] = Integer.MAX_VALUE;
          }
          for(int i = 1; i * i <= n; i++){
              dp[i * i] = 1;
          }
          for(int i = 2; i <= n; i++){
              if(dp[i] == 1) continue;
              for(int j = i / 2; j >= 1; j--){
                  dp[i] = Math.min(dp[i], dp[j] + dp[i - j]);
              }
          }
          return dp[n];
      }
  }

• Method 2 dp

class Solution {
    public int numSquares(int n) {
        int[] dp = new int[n + 1];
        for(int i = 1; i <= n; i++){
            int res = Integer.MAX_VALUE;
            for(int j = 1; j * j <= i; j++)
                res = Math.min(res, dp[i - j * j] + 1);
            dp[i] = res;
        }
        return dp[n];
    }
}