LeetCode Perfect Squares

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：

Let dp[i] denotes the least number of perfect squares sum to i.

Then for all the candiates smaller than i, if the difference between i and candidate is perfect square, then update the dp[candidate]+1. Maintain the smallest.

Thne how to make sure the difference is perfect square.

Let candidate = i - j*j.

Time Complexity: O(nlogn).

Space: O(n).

AC Java:

 class Solution {
     public int numSquares(int n) {
         int [] dp = new int[n+1];
         for(int i = 1; i<=n; i++){
             dp[i] = i;
             for(int j = 0; i-j*j>=0; j++){
                 dp[i] = Math.min(dp[i], dp[i-j*j]+1);
             }
         }

         return dp[n];
     }
 }

Could also do it from head to tail.

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:

 public class Solution {
     public int numSquares(int n) {
         if(n < 0){
             return 0;
         }
         int [] dp = new int[n+1];
         Arrays.fill(dp, Integer.MAX_VALUE);
         for(int i = 0; i*i <= n; i++){
             dp[i*i] = 1;
         }
         for(int i = 1; i<=n; i++){
             for(int j = 1; i+j*j<=n; j++){
                 dp[i+j*j] = Math.min(dp[i+j*j], dp[i]+1);
             }
         }
         return dp[n];
     }
 }

与Count Primes类似.