【LeetCode】455. Assign Cookies

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本文探讨了一个经典的分配问题——如何将有限大小的饼干分配给具有不同需求的孩子们，以最大化满足孩子的数量。通过使用贪心算法，我们对孩子们的需求和饼干的大小进行排序，然后尝试将每个饼干分配给最能接受它的孩子。文章提供了详细的算法实现，并通过两个实例展示了其工作原理。

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455. Assign Cookies

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：贪心选择

#include<iostream>
#include<algorithm>
#include<vector> 
using namespace std;
int findContentChildren(vector<int>& g, vector<int>& s) {
    sort(g.begin(),g.end());
    sort(s.begin(),s.end());
    int ans=0;
	for(int gi=0,si=0;gi<g.size()&&si<=s.size();){
		if(g[gi]<=s[si]) {
			ans++;
			gi++;	//如果可以满足，则应该开始满足下一个要吃蛋糕的小朋友了 
		}
		si++;//如果可以满足，换到下一个蛋糕；
			//如果不可以满足，说明当前这个蛋糕谁都满足不了（因为这是排序过了的），只能放弃 
	}
	return ans;
}
 int main(void)
 {
 	vector<int> g,s;
 	int g_size,s_size;
 	while(cin>>g_size>>s_size){
 		int num;
 		for(int i=0;i<g_size;i++){
 			cin>>num;
 			g.push_back(num);
		 }
		for(int i=0;i<s_size;i++){
 			cin>>num;
 			s.push_back(num);
		 }
		 cout<<findContentChildren(g,s)<<endl;
	 }
	 return 0;
 }