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.
Example 1:
Input: g = [1,2,3], s = [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: g = [1,2], s = [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.
Constraints:
1 <= g.length <= 3 * 1040 <= s.length <= 3 * 1041 <= g[i], s[j] <= 231 - 1
基础贪心。
题意为有m个饼干每个饼干大小为s[i]，有n个孩子每个孩子有一个贪婪因数为g[j]。给孩子们发饼干，第j个孩子只接受s[i]>=g[j]的饼干i，若得到符合条件的饼干，孩子得到满足。给定s[]和g[]，求出能满足孩子的最大数量。
本题用贪心思想解决，即每次给当前满足因数最小的孩子他所能满足的最小的饼干。
详细点展开一步。将饼干按大小由小到大分发，对于每个饼干s[i]，如果存在能以其满足的孩子，则发给能够满足的满足因数g[j]最小的孩子，如果不存在能以其满足的孩子，则说明这块饼干太小不能满足任何孩子，指针i后移到下一个饼干s[i+1]，执行重复操作，直到遍历完成所有饼干。
再详细点展开一步。将饼干按大小由小到大分发，对于每个饼干s[i]，按顺序与同为升序排列的孩子贪婪因数数组各项g[j]对比，找到满足s[i]>=g[j]的第一个j，将第i个饼干发给第j个孩子，计数后使饼干指针和孩子指针都后移判断下一个饼干i+1和孩子j+1，如果并没有g[j]满足s[i]>=g[j]，则说明当前饼干无法满足任何一个孩子，使饼干指针后移判断下一个饼干i+1，直到遍历完成所有饼干。
