279. Perfect Squares(动态规划)

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.

dp[0] = 0
dp[1] = dp[0]+1 = 1
dp[2] = dp[1]+1 = 2
dp[3] = dp[2]+1 = 3
dp[4] = Min{ dp[4-1*1]+1, dp[4-2*2]+1 } = Min{ dp[3]+1, dp[0]+1 } = 1
dp[5] = Min{ dp[5-1*1]+1, dp[5-2*2]+1 } = Min{ dp[4]+1, dp[1]+1 } = 2
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dp[13] = Min{ dp[13-1*1]+1, dp[13-2*2]+1, dp[13-3*3]+1 } = Min{ dp[12]+1, dp[9]+1, dp[4]+1 } = 2
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dp[n] = Min{ dp[n - i*i] + 1 }, n - i*i >=0 && i >= 1

 class Solution {
 public:
     int numSquares(int n) {
         vector<int> dp(n+,);
         for(int i = ;i<=n;i++){

             int mins = INT_MAX;
             for(int j = ;i-j*j>=;j++){
                 mins = min(mins,dp[i-j*j]+);
             }
             dp[i] = mins;
         }
         return dp[n];
     }
 };