题目:
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.
分析:DP问题,设dp[n]为输入正整数n时,函数返回的结果。找规律如下:
dp[0] = 0 dp[1] = 1 dp[2] = 2 dp[3] = 3 dp[4] = 1 = dp[0] + 1 = dp[4-2*2] + 1 dp[5] = 2 dp[6] = 3 dp[7] = 4 dp[8] = 2 = dp[4] + 1 = dp[8-2*2] + 1 dp[9] = 1 = dp[0] + 1 = dp[9-3*3] + 1
代码如下:
public int numSquares(int n) { int[] dp = new int[n+1]; dp[0] = 0; for(int i = 1; i <= n; i++) { dp[i] = dp[i-1]+1; for(int j = 2; j*j <= i; j++) { dp[i] = Math.min(dp[i], dp[i-j*j]+1); } } return dp[n]; }
原文:http://www.cnblogs.com/lasclocker/p/5042323.html