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.

 

这道题给了我们一堆cookie，每个cookie的大小不同，还有一堆children，每个小朋友的胃口不同，问我们当前的cookie最多能满足几个小朋友。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。

 

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。

 

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;

    }

};

 

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