2017算法课.08(Assign Cookies )

455. Assign Cookies

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• Total Accepted: 21109
• Total Submissions: 44746
• Difficulty: Easy
• Contributors: love_Fawn

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 g_i, which is the minimum size of a cookie that the child will be content with; and each cookie j has a size s_j. If s_j >= g_i, 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: 1Explanation: 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: 2Explanation: 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.

题目要求：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：

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;    }};