LeetCode -- Assign Cookies

问题描述:

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.

输入是两个数组，分别为children所需最小的cookie大小，和cookie数组。任务是需要将cookie分配到children手中，满足cookie[i]>=children[j]。求可满足最多children的数量。

思路：
贪心策略：
children数组排序，cookie排序。
两个索引(childrenIndex, cookieIndex)来遍历children和cookie数组。
找到可满足children的最小cookie，满足后childrenIndex++,cookieIndex+1后继续遍历，直到cookie分配完或children都被分配为止。
返回children的索引作为结果。

实现代码：
public class Solution {
public int FindContentChildren(int[] g, int[] s) {
if(s == null || s.Length == 0){
return 0;
}

g = g.OrderBy(x=>x).ToArray();
s = s.OrderBy(x=>x).ToArray();

var childIndex = 0;
var cookieIndex = 0;
while(childIndex < g.Length && cookieIndex < s.Length){
// get the minimum cookie index
var start = cookieIndex;
while (start < s.Length && g[childIndex] > s[start])
{
start ++;
}
cookieIndex = start ;
//Console.WriteLine(cookieIndex);
if(cookieIndex < s.Length){
childIndex ++;
}
cookieIndex ++;
}

return childIndex;
}
}