455. Assign Cookies

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最新推荐文章于 2025-01-17 23:32:35 发布

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解决如何最大数量地满足孩子们的需求，通过合理分配不同大小的饼干。该问题需遍历排序后的饼干和孩子需求数组，确保每个孩子最多只能获得一块饼干。

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题目：

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.

题解：

分配饼干尽量以饼干尺寸最小并且能满足小孩的要求。

先将饼干尺寸和小孩需求都排个序，然后从小到大去循环遍历两个数组。遍历过程中有两个条件判断：

（1）如果饼干大小不能满足小孩，则饼干数组就往后移一个下标位置；

（2）如果饼干大小大于小孩的greed factor g，则孩子数组和饼干数组同时往后移一个下标位置，且被满足的孩子个数plus 1；

循环的结束条件就是小孩或者饼干的标记二者有一个到了数组的尾部，就算停止。

代码如下：

class Solution {
public:
    int findContentChildren(vector<int>& g, vector<int>& s) {
        sort(g.begin(),g.end());
        sort(s.begin(),s.end());
        
        int num = 0;
        int child = 0;
        int cookie = 0;
        while(child < g.size() && cookie < s.size()) {
            if(g[child] > s[cookie])
                cookie++;
            else {
                child++;
                cookie++;
                num++;
            } 
        }
        return num;
    }
};

end!