Leetcode刷题96-455. 分发饼干(C++详细解法！！！)

Come from : [https://leetcode-cn.com/problems/assign-cookies/]

1.Question

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.

2.Answer

easy 类型题目。.贪心算法。
算法思路：

1. 现将两个数组g和s，从小到大排序。
2. 按照从小到大的顺序使用各个糖果看是否满足孩子的需求，每个糖果只尝试一次（贪心算法）。
3. 若尝试成功，则换下一个孩子，直到发现 没有更多的孩子或者没有更多的糖果，循环结束。

AC代码如下：

class Solution {
public:
    int findContentChildren(vector<int>& g, vector<int>& s) {
        sort(g.begin(), g.end());  //排序
        sort(s.begin(), s.end());
        int child = 0;  //定义两个辅助指针变量
        int cookie = 0;
        while(child < g.size() && cookie < s.size()) //循环结束条件
        {
            if(g[child] <= s[cookie]) //若糖果满足还租的需求，孩纸对应的指针后移1个位置
            {
                ++child;
            }
            ++cookie;  //否则，糖果对应的指针后移1个位置
        }
        return child;
    }
};

3.我的收获

贪心算法初步。。。
fighting。。。
端午节快乐~记得吃粽子。。。

2019/6/7 胡云层 于南京 96