原题链接在这里:https://leetcode.com/problems/perfect-squares/
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
.
题解:
与Count Primes类似. 初始化i*i的位置为1, 然后对i + j*j 更新min(dp[i+j*j], dp[i] + 1).
Time Complexity: O(nlogn). Space: O(n).
AC Java:
1 public class Solution { 2 public int numSquares(int n) { 3 if(n < 0){ 4 return 0; 5 } 6 int [] dp = new int[n+1]; 7 Arrays.fill(dp, Integer.MAX_VALUE); 8 for(int i = 0; i*i <= n; i++){ 9 dp[i*i] = 1; 10 } 11 for(int i = 1; i<=n; i++){ 12 for(int j = 1; i+j*j<=n; j++){ 13 dp[i+j*j] = Math.min(dp[i+j*j], dp[i]+1); 14 } 15 } 16 return dp[n]; 17 } 18 }