LeetCode 279. Perfect Squares

Question:

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.

 

Code:

public class Solution {
    public int numSquares(int n) {
        Queue<Map.Entry<Integer, Integer>> queue = new LinkedList<>();
        queue.add(new java.util.AbstractMap.SimpleEntry<Integer, Integer>(n, 0));

        boolean[] visited = new boolean[n+1];
        visited[n] = true;
        while (!queue.isEmpty()){
            Map.Entry<Integer, Integer> entry = queue.poll();
            int num = entry.getKey();
            int step = entry.getValue();

            for (int i = 1; ; i++) {
                int a = num - i*i;
                if(a < 0){
                    break;
                }
                if(a == 0){
                    return step + 1;
                }
                if (!visited[a]){
                    queue.add(new java.util.AbstractMap.SimpleEntry<Integer, Integer>(a, step + 1));
                    visited[a] = true;
                }
            }

        }
        return 0;
    }
}

 

动态规划

class Solution {
    private boolean isSquares(int n){
        int temp = (int)Math.sqrt(n);
        return temp*temp == n ? true:false;
    }
    public int numSquares(int n){
        if(isSquares(n)){
            return 1;//如果是平方数,直接返回1
        }
        int[] dp = new int[n+1];
        for(int i = 0; i < dp.length; i++){
            if(isSquares(i)){
                dp[i] = 1;
            }else{
                dp[i] = Integer.MAX_VALUE;
            }
        }
        for (int i = 2; i < n + 1; i++) {
            for (int j = 1; j*j < i; j++) {
                dp[i] = Math.min(dp[i], dp[j*j]+dp[i-j*j]);
            }
        }
        return dp[n];
    }
}

 

posted @ 2017-07-27 21:23  Realvie  阅读(116)  评论(0编辑  收藏  举报