LeetCode 279. Perfect Squares

最新推荐文章于 2025-09-06 09:33:29 发布

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原文链接：http://www.cnblogs.com/realvie/p/7247454.html

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#java #数据结构与算法

本文介绍了一种算法，用于找出组成正整数n所需的最少完全平方数的数量。通过两种方法实现：一种采用广度优先搜索策略；另一种使用动态规划。前者在代码中利用队列和哈希表记录状态；后者则通过构建一维数组来迭代计算最小值。

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];
    }
}