leetCode---Assign Cookies

一. 题目：Assign Cookies   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 andoutput 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.

二. 思路分析

题目大意：有一堆饼干，每个饼干的大小不同，还有一堆小朋友，每个小朋友的胃口也不同的，问我们当前的饼干最多能满足几个小朋友。

 
      思路分析：这是典型的利用贪婪算法的题目，我们可以首先对两个数组进行排序，让小的在前面。然后我们先拿最小的cookie给胃口最小的小朋友，看能否满足，能的话，我们结果res自加1，然后再拿下一个cookie去满足下一位小朋友；如果当前cookie不能满足当前小朋友，那么我们就用下一块稍大一点的cookie去尝试满足当前的小朋友。当cookie发完了或者小朋友没有了我们停止遍历，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 i = g.size() - 1, j = s.size() - 1, count = 0;
while(i >= 0 && j >= 0){
if(g[i] > s[i]) i--;
else if(g[i--] <= s[j--]) count++;
}
return count;
}
};