LeetCode：Assign Cookies

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.

Example1：

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.

Example2：

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.

思路：

输入：两个数组，分别表示能满足小孩子的cookie的最小尺寸、cookies的不同尺寸。

要求：分给每个小孩子一个cookie，最大可能满足他们。

输出：最多可以满足几个小孩子。

　　这是一道简单的利用贪婪算法的题，需要尽可能的把小cookie发给greed factor小的小孩子。可以先对两个数组进行升序排序，然后从greed factor最小的小孩子以及最小的cookie开始，如果该cookie满足该小孩子，则结果加1，拿下一块cookie去满足下一个小孩子；否则，继续拿下一块cookie来满足当前小孩子。一直到cookies发完了或者没有小孩子才停止遍历。

代码：

class Solution {
public:
int findContentChildren(vector<int>& g, vector<int>& s) {
sort(g.begin(), g.end());
sort(s.begin(), s.end());
int i = 0;

4000
int j = 0;
while (i < g.size() && j < s.size()) {
if (s[j] >= g[i]) {
i++;
j++;
} else {
j++;
}
}
return i;
}
};

时间复杂度分析：

排序需要的时间为O(nlogn)，遍历需要的时间为O(max{g.size(), s.size()})

对于输入规模为n的问题，总的时间复杂度为O(n+nlogn) = O(nlogn)