【LeetCode】455. Assign Cookies (java实现)
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【LeetCode】455. Assign Cookies (java实现)
BookShu 发表于1年前
【LeetCode】455. Assign Cookies (java实现)
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腾讯云 技术升级10大核心产品年终让利>>>   

摘要: Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie. Each child i has a greed factor gi, which is the minimum size of a cookie that the child will be content with; and each cookie j has a size sj. If sj >= gi, we can assign

原题链接

https://leetcode.com/problems/assign-cookies/

原题

Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie. Each child i has a greed factor gi, which is the minimum size of a cookie that the child will be content with; and each cookie j has a size sj. If sj >= gi, we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.

Note: You may assume the greed factor is always positive. You cannot assign more than one cookie to one child.

Example 1:

Input: [1,2,3], [1,1]

Output: 1

Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3. 
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1.

Example 2:

Input: [1,2], [1,2,3]

Output: 2

Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. 
You have 3 cookies and their sizes are big enough to gratify all of the children, 
You need to output 2.

题目要求

题目叫“分配饼干”,给定两个数组,分别表示每个小孩期望的饼干尺寸,和每个饼干实际的尺寸。将饼干分配给这些小孩,但分配的饼干尺寸必须不小于小孩期望的饼干尺寸。求出这些饼干最多可以满足几个小孩。

解法

解法:题目比较简单清晰,既然是饼干尺寸不小于期望尺寸,那么我们首先将两个数组先排序。遍历期望尺寸的数组和饼干尺寸的数组,如果饼干尺寸符合则两个数组都向前进一,表示有一个饼干满足了一个小孩;如果饼干尺寸不符合,则饼干数组向前进一,尝试下一个饼干。

public int findContentChildren(int[] g, int[] s) {
    int ret = 0;
    
    Arrays.sort(g);
    Arrays.sort(s);
    
    int i = 0, j = 0;
    while (i < g.length && j < s.length) {
        if (g[i] <= s[j]) {
            ret++;
            i++;
            j++;
        }else if (g[i] > s[j]) {
            j++;
        }
    }
    
    return ret;
}

测试用例:

public static void main(String[] args) {
    Solution so = new Solution();
    {
        int []g = {1, 2, 3};
        int []s = {1, 1};
        assert(so.findContentChildren(g, s) == 1);
    }
    {
        int []g = {1, 2};
        int []s = {1, 2, 3};
        assert(so.findContentChildren(g, s) == 2);
    }
    {
        int []g = {};
        int []s = {};
        assert(so.findContentChildren(g, s) == 0);
    }
}
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