279. Perfect Squares
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Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
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题解：
四平方和定理这种可遇不可求的做法就不说了，说说dp:
dp[i]代表组成i的最少平方数的个数。至于状态转移，我们可以这样想：组成这个数的最小值要么就是本身，要么就是前面某一个数+一个平方数（所以看作值加上1）,类似0-1背包的思想。
Submission Details
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600 / 600 test cases passed.
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Status:
Accepted |
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Runtime: 461 ms
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1 class Solution { 2 public: 3 int numSquares(int n) { 4 vector<int> dp(n+1,INT_MAX); 5 dp[0] = 0; 6 dp[1] = 1; 7 for(int i=2; i<=n; i++) { 8 for(int j = 1;j * j <= i ;j++){ 9 dp[i] = min(dp[i], dp[i - j * j] + 1); 10 } 11 } 12 return dp[n]; 13 } 14 };