[Leetcode] 455. Assign Cookies 解题报告
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题目:
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: 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.
思路:
注意到能给某个孩子分配饼干的充要条件是S_j >= G_i。所以最优方法就是,首先对S和G都进行排序,然后看看当前的S_j能不能分配给G_i,如果可以,就分配(这一定是最优选择),否则就试着给他分配S_(j+1)...直到饼干被试完(分完)。算法的时间复杂度是O(nlogn)+O(mlogm),其中n和m分别是S和G的长度。
代码:
class Solution {public: int findContentChildren(vector<int>& g, vector<int>& s) { sort(g.begin(), g.end()); sort(s.begin(), s.end()); int i = 0, j = 0, ret = 0; while (i < g.size() && j < s.size()) { while (j < s.size() && s[j] < g[i]) { ++j; } if (j < s.size() && s[j] >= g[i]) { ++ret; ++j, ++i; } } return ret; }};
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