leetcode assign-cookies

Posted 2021-02-08 zhibin123

455. Assign Cookies

Easy

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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;
    }
};