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
分发饼干。
假设你是一位很棒的家长，想要给你的孩子们一些小饼干。但是，每个孩子最多只能给一块饼干。
对每个孩子 i，都有一个胃口值 g[i]，这是能让孩子们满足胃口的饼干的最小尺寸；并且每块饼干 j，都有一个尺寸 s[j] 。如果 s[j] >= g[i]，我们可以将这个饼干 j 分配给孩子 i ，这个孩子会得到满足。你的目标是尽可能满足越多数量的孩子，并输出这个最大数值。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/assign-cookies
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思路是two pointer。首先把两个input数组排序，这样需求最小的孩子在前，尺寸最小的饼干也在前。用双指针的方式遍历这两个数组，如果当前饼干能满足当前孩子的需求，则知道有一个孩子的需求被满足了，否则就看下一个饼干是否能满足其需求。饼干遍历完毕的时候，看看有几个孩子的需求能被满足。
时间O(nlogn)
空间O(1)
Java实现
1 class Solution { 2 public int findContentChildren(int[] g, int[] s) { 3 Arrays.sort(g); 4 Arrays.sort(s); 5 int i = 0; 6 int j = 0; 7 while (i < g.length && j < s.length) { 8 if (g[i] <= s[j]) { 9 i++; 10 j++; 11 } else { 12 j++; 13 } 14 } 15 return i; 16 } 17 }