#week16
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
这一题看上去有点晕的样子
但是实际上也是一道简单的贪心
每个孩子有g[i]的曲奇
有s数组的曲奇
想让得到满足的孩子的数量尽可能多
因此将两个数组排个序
都从最大的开始分
分别比较
能够分到的就分
不能就往下一个孩子
然后计数一下就可以得到答案
题解
1 class Solution { 2 public: 3 int findContentChildren(vector<int>& g, vector<int>& s) { 4 sort(g.begin(),g.end()); 5 sort(s.begin(),s.end()); 6 int i=g.size()-1, j=s.size()-1,count = 0; 7 while(i>=0 && j>=0) 8 { 9 if(g[i]>s[j]) i--; 10 else if(g[i--]<=s[j--]) count++; 11 } 12 return count; 13 } 14 };