标签：
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.
解法一：
dp[n] indicates that the perfect squares count of the given n, and we have:
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 . . . 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 . . . dp[n] = Min{ dp[n - i*i] + 1 }, n - i*i >=0 && i >= 1
public int numSquares(int n) { int[] dp = new int[n + 1]; Arrays.fill(dp, Integer.MAX_VALUE); dp[0] = 0; for(int i = 1; i <= n; ++i) { int min = Integer.MAX_VALUE; int j = 1; while(i - j*j >= 0) { min = Math.min(min, dp[i - j*j] + 1); ++j; } dp[i] = min; } return dp[n]; }
reference：
https://discuss.leetcode.com/topic/26400/an-easy-understanding-dp-solution-in-java/2
解法二：faster 的DP
https://discuss.leetcode.com/topic/24255/summary-of-4-different-solutions-bfs-dp-static-dp-and-mathematics/25
标签：
原文地址：http://www.cnblogs.com/hygeia/p/5751641.html
