455. Assign Cookies
Easy
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.
题意:
给了孩子的贪婪参数,饼干的大小,求可以满足最多多少孩子。
思路:
最节约的分给孩子饼干就可以得到最后最多。贪心。
代码:
class Solution { public: int findContentChildren(vector<int>& g, vector<int>& s) { int ans = 0; sort(g.begin(),g.end()); sort(s.begin(),s.end());//排好序 for(int i = 0;i < g.size();i++){//遍历孩子,找到最低能满足要求的饼干给他,并标记饼干已经分出,分给后找下一个孩子 for(int j = 0;j < s.size();j++){ if(g[i] <= s[j]){ s[j] = 0; ans++; break; } } } return ans; } };
改进:
每次找到合适的饼干后,之后的孩子只能得到尺寸不小于当前的饼干,所以,j=0放前面就好。
class Solution { public: int findContentChildren(vector<int>& g, vector<int>& s) { int ans = 0; sort(g.begin(),g.end()); sort(s.begin(),s.end()); int j = 0; for(int i = 0;i < g.size();i++){ for(;j < s.size();j++){ if(g[i] <= s[j]){ s[j] = 0; ans++; break; } } } return ans; } };