[LeetCode] Assign Cookies 分点心

2023年3月1日 184次阅读 来源: Grandyang

 

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.

 

这道题给了我们一堆cookie，每个cookie的大小不同，还有一堆小朋友，每个小朋友的胃口也不同的，问我们当前的cookie最多能满足几个小朋友。这是典型的利用贪婪算法的题目，我们可以首先对两个数组进行排序，让小的在前面。然后我们先拿最小的cookie给胃口最小的小朋友，看能否满足，能的话，我们结果res自加1，然后再拿下一个cookie去满足下一位小朋友；如果当前cookie不能满足当前小朋友，那么我们就用下一块稍大一点的cookie去尝试满足当前的小朋友。当cookie发完了或者小朋友没有了我们停止遍历，参见代码如下：

 

解法一：

class Solution {
public:
    int findContentChildren(vector<int>& g, vector<int>& s) {
         int res = 0, p = 0;
         sort(g.begin(), g.end());
         sort(s.begin(), s.end());
         for (int i = 0; i < s.size(); ++i) {
             if (s[i] >= g[p]) {
                 ++res;
                 ++p;
                 if (p >= g.size()) break;
             }
         }
         return res;
    }
};

 

我们可以对上述代码进行精简，我们用变量j既可以表示小朋友数组的坐标，同时又可以表示已满足的小朋友的个数，因为只有满足了当前的小朋友，才会去满足下一个胃口较大的小朋友，参见代码如下：

 

解法二：

class Solution {
public:
    int findContentChildren(vector<int>& g, vector<int>& s) {
        int j = 0;
        sort(g.begin(), g.end());
        sort(s.begin(), s.end());
        for (int i = 0; i < s.size() && j < g.size(); ++i) {
            if (s[i] >= g[j]) ++j;
        }
        return j;
    }
};

 

参考资料：

https://discuss.leetcode.com/topic/67676/simple-greedy-java-solution

 

    原文作者：Grandyang
    原文地址: http://www.cnblogs.com/grandyang/p/6077344.html
    本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。
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