[Leetcode] 455. Assign Cookies 解题报告

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.

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.

思路：

注意到能给某个孩子分配饼干的充要条件是S_j >= G_i。所以最优方法就是，首先对S和G都进行排序，然后看看当前的S_j能不能分配给G_i，如果可以，就分配（这一定是最优选择），否则就试着给他分配S_(j+1)...直到饼干被试完（分完）。算法的时间复杂度是O(nlogn)+O(mlogm)，其中n和m分别是S和G的长度。

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