Leetcode:455. Assign Cookies (week 7)

Description:

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.

解题思路

　本题意味分发曲奇给尽可能多的小朋友，并且曲奇饼干的大小必须满足小朋友的胃口。因此不能把大的曲奇去满足很小胃口的小朋友，尽可能的去把小曲奇发给小胃口的小朋友，使得尽可能多的小朋友分到。本题涉及贪心算法，要满足尽可能多的小孩，解题思路也比较简单（如下）。
　算法思路：
　1.将饼干尺寸和小孩需求都排个序【可应用C++中改进的快排算法sort()函数，时间复杂度为O(nlogn)】
　2.从小到大去遍历地给，在遍历过程中可在两个排序后的数组都设一个标记，一起往后移，饼干大小不满足就移饼干的标记，看看后面的饼干能不能满足他，只有满足了才移小孩的标记，因为如果这个小孩都不能满足，后面更贪婪的小孩更加满足不了。
　3. 循环的结束条件为小孩或者饼干的标记二者有一个到了数组的尾部。
　总的时间复杂度为O(nlogn)

代码

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