【算法作业9】LeetCode 455. Assign Cookies

122. Best Time to Buy and Sell Stock II

Description:

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: 1Explanation: 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: 2Explanation: 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.

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题解：

此题是一道简单的贪心算法的题目。题目的输入为两个vector，第一个vector代表每个孩子的greed factor，第二个vector代表每块饼干的大小。只有当孩子分到的饼干的大小比greed factor大，孩子才会感到满足。要求输出最多能有多少个孩子感到满足。

对于本题，我的思路是，先让两个vector按照由小到大进行排序，然后用两层for循环逐个从小到大进行比较，因为要保证先把能够满足一个孩子的最小的饼干给他。每次找到一个最小size又能让孩子满足的一个饼干，就把这个饼干给他，count递增1，并且将此饼干的size置为0（因为孩子的greed factor恒大于零，这样做相当于饼干已经被拿走了），跳出内层循环。最后得到count的数目就是最多能有多少个孩子感到满足。

代码：

#include <algorithm>class Solution {public:    int findContentChildren(vector<int>& g, vector<int>& s) {        sort(g.begin(), g.end());        sort(s.begin(), s.end());        int count = 0;        for (int i = 0; i < g.size(); i++)        {            for (int j = 0; j < s.size(); j++)            {                if (g[i] <= s[j])                {                    count++;                    s[j] = 0;                    break;                }            }        }        return count;    }};