Leetcode: 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.

Solution 1: Greedy, Time:O(nlogn)

Just assign the cookies starting from the child with less greediness to maximize the number of happy children .

 1 public class Solution {
 2     public int findContentChildren(int[] g, int[] s) {
 3         Arrays.sort(g);
 4         Arrays.sort(s);
 5         int i = 0;
 6         for (int j=0; i<g.length && j<s.length; j++) {
 7             if (g[i] <= s[j]) i++;
 8         }
 9         return i;
10     }
11 }

Solution 2: Greedy + TreeMap,   Time: O(nlogm), n, m are the size of two arrays respectively

 1 public class Solution {
 2     public int findContentChildren(int[] g, int[] s) {
 3         TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
 4         int res = 0;
 5         for (int size : s) {
 6             map.put(size, map.getOrDefault(size, 0)+1);
 7         }
 8 
 9         for (int greed : g) {
10             if (map.ceilingKey(greed) != null) {
11                 res++;
12                 int value = map.ceilingKey(greed);
13                 map.put(value, map.get(value)-1);
14                 if (map.get(value) == 0) map.remove(value);
15             }
16         }
17         return res;
18     }
19 }
原文地址:https://www.cnblogs.com/EdwardLiu/p/6168415.html