LeetCode 455. Assign Cookies

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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.

分析:先将g数组和s数组从小到大排序~ i指针遍历g数组,j指针遍历s数组,如果当前g[i] <= s[j],也就是当前糖果j能够分给当前小朋友i,那就分配,并且将i指针指向下一个小朋友,cnt同时也要累加一个~如果当前糖果不能分配给当前小朋友,说明糖果j不能分给任何人了,因为没有比小朋友i需要的还要少的小朋友了(i后面的数都比i大)。所以无论是否分配,都把j向后移动一次~这样能保证需求量小的所有小朋友都分配得到能够分配的糖果,此时的cnt也是所求的贪心最大值~避免了大材小用(大糖果分配给需求量小的小朋友)的情况~

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

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