1. 题目描述
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.
2. 样例
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.
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.
3. 分析
【理解】：给了两个输入的数组，分别代表了每个孩子期待分到的cookie的大小和所拥有的cookie的大小情况。因此第一个数组的长度就是孩子的总数，第二个数组的长度就是cookie的总数。我们需要求出怎么样分配：才能使最多数量的孩子对cookie的需求得到满足。
【思路】：我个人觉得用贪心也比较容易想思路：我们将所有的cookie按照大小降序排序，将所有的孩子期待得到cookie的大小也降序排序，接下来两个数组同时开始遍历，这样的做法能够保证，每一步骤下都可以判断当前的孩子是否可以满足心愿，因为当前的cookie也是最大的（比其更大的cookie已经分给了需求更大的孩子）。
如果当前cookie大小大于等于当前孩子的需求：说明这块cookie是可以满足条件的，记录下来；如果当前cookie大小小于当前孩子的需求：那么说明这个孩子的需求得不到满足，就只好舍弃他，我们去访问下一个孩子看看能否满足他的需求，以求更多的孩子能分到cookie。
4. 源码
class Solution {
public:
static bool Comp(int a, int b) {
return (a > b);
}
int findContentChildren(vector<int>& g, vector<int>& s) {
sort(g.begin(), g.end(), Comp);
sort(s.begin(), s.end(), Comp);
int counter = 0;
int i = 0, j = 0;
while(i < (int)s.size() && j < (int)g.size()) {
if (s[i] >= g[j]) {
counter++;
i++;
j++;
}
else {
j++;
}
}
return counter;
}
};5. 心得
这道题属于典型的贪心算法的最简单应用，之所以简单是每个孩子最多只允许被分配一个cookie，如果cookie可以累加，那么问题就复杂了很多。