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.
分析:
有两个数组,分别代表孩子需要曲奇的大小,和你所拥有曲奇的大小,求最大满足孩子的数目。
经典的贪心问题,我们可以这样想,如果一个曲奇不能满足当前这个孩子的话,那么一定不能满足比这个孩子还要大的尺寸。我们还希望满足一个孩子曲奇的尺寸尽量和他需要的接近,也就是不要浪费大尺寸的曲奇。所以我们可以先将两个数组有小到大排序。遍历曲奇数组,如果当前这个曲奇不满足第一个孩子,那么就看下一个曲奇。如果满足,则数目加一,并且孩子和曲奇的指针也同时加一。这样可以满足我们上面提到的两个要求。
程序:
class Solution { public: int findContentChildren(vector<int>& g, vector<int>& s) { sort(g.begin(), g.end()); sort(s.begin(), s.end()); int sum = 0; int i = 0; int j = 0; while(i < g.size() && j < s.size()){ while(j < s.size()){ if(g[i] <= s[j]){ sum++; j++; break; } j++; } i++; } return sum; } };