Problem:
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.
Summary:
将M个饼干分给N个小朋友,每个小朋友有一个贪心值,要求小朋友的得到的饼干大小要大于或等于贪心值大小。
求最多有多少小朋友可以分到饼干。
Solution:
将两个数组由小到大排序,分别用指针从头至尾比较即可。
1 class Solution { 2 public: 3 int findContentChildren(vector<int>& g, vector<int>& s) { 4 int cnt = 0, glen = g.size(), slen = s.size(); 5 sort(g.begin(), g.end()); 6 sort(s.begin(), s.end()); 7 8 int i = 0, j = 0; 9 while ( i < glen && j < slen) { 10 if (g[i] <= s[j]) { 11 i++; 12 j++; 13 cnt++; 14 } 15 else if (g[i] > s[j]) { 16 j++; 17 } 18 } 19 20 return cnt; 21 } 22 };