LeetCode #455: Assign Cookies

Problem

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.

Solution

题意

• 输入有两个数组g和s
• 第一个数组代表有g.size()个孩子，每个孩子i的“贪心因数”是g[i]
• 第二个数组代表现有s.size()块饼干，每块饼干i的大小是s[i]
• 对于每个孩子，必须要满足：分到的饼干的大小要大于等于该孩子的“贪心因数”，才能称为“该孩子得到了满足”
• 对于每个孩子，只能分不多于一块的饼干

分析

这是一道贪心的题目，最最自然的思路就是：
1. 首先对两个数组分别进行排序
2. 同时开始遍历两个数组：对于孩子g[i]，如果有饼干s[j]能满足他，那么i++，如果不能，如果不能，则j++，判断下一块饼干是否能满足他
3. 循环的终止条件是i < g.size() && j < s.size()

Code

class Solution {public:    int findContentChildren(vector<int>& g, vector<int>& s) {        sort(g.begin(), g.end());        sort(s.begin(), s.end());        int i = 0;        int j = 0;        for (; i < g.size() && j < s.size(); j++){            if (g[i] <= s[j])                i++;        }        return i;    }};//Runtime: 46ms

复杂度分析

• 时间复杂度：排序需要的时间为O(nlogn)，遍历需要的时间为O(max{g.size(), s.size()})；对于输入规模为n的问题，总的时间复杂度应该是O(n+nlogn) = O(nlogn)
• 空间复杂度：O(n)