leetcode 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: 1Explanation: 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: 2Explanation: 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.

这一题跟51nod抓兔子那一题有所类似，但是更简单，根本就用不上优先级队列。

package leetcode;public class Assign_Cookies_455 {public int findContentChildren(int[] g, int[] s) {if(g.length==0||s.length==0){return 0;}quicksort(g, 0, g.length-1);//childrenquicksort(s, 0, s.length-1);//cookiesint cookiePointer=0;for(int chidrenPointer=0;chidrenPointer<g.length;chidrenPointer++){int childrenNeed=g[chidrenPointer];if(childrenNeed>s[cookiePointer]){continue;}else{cookiePointer++;if(cookiePointer==s.length){break;}}}return cookiePointer;}//倒序快排public void quicksort(int[] a,int left,int right){if(left<right){int low=left;int high=right;int pivot=a[low];while(low<high){while(low<high&&a[high]<=pivot){high--;}if(low<high){a[low]=a[high];low++;}while(low<high&&a[low]>=pivot){low++;}if(low<high){a[high]=a[low];high--;}}a[low]=pivot;quicksort(a, left, low-1);quicksort(a, low+1, right);}}public static void main(String[] args) {// TODO Auto-generated method stubAssign_Cookies_455 a=new Assign_Cookies_455();int[] g=new int[]{1,2,3};int[] s=new int[]{3};System.out.println(a.findContentChildren(g, s));}}

大神想法跟我类似，只不过是从小到大排序，然后搜索第一个比孩子欲望大的饼。

public class Solution {    public int findContentChildren(int[] g, int[] s) {        Arrays.sort(g);        Arrays.sort(s);                int pointG = 0;        int pointS = 0;                while (pointG<g.length && pointS<s.length) {            if (g[pointG]<=s[pointS]) {                pointG++;                pointS++;            } else {                pointS++;            }        }                return pointG;    }}

另外一种方法是，使用treemap，treemap内部实现原理是红黑树哦。

首先用treemap记录cookie的<size,个数>，treemap有一个方法叫ceilingKey，The method call returns the least key greater than or equal to key, or null if there is no such key.再对于每个children的贪婪度进行遍历，如果有正好大于等于它的饼，那么就对map做出修改：如果该size只有一个，那么个数减少后，个数为0，那么从map中去掉这个size；如果个数不为1，那么个数--。

public class AssignCookies {    public static int findContentChildren(int[] g, int[] s) {    int count = 0;    TreeMap<Integer,Integer> tree = new TreeMap<>();    for(int temp : s){    Integer num = tree.get(temp);    num = num==null?0:num;    tree.put(temp,num+1);    }    for(int temp : g){    Integer targ = tree.ceilingKey(temp);    if(targ!=null){    Integer num = tree.get(targ);    if(num>0){    count++;    if(num==1){    tree.remove(targ);    }else{                                        tree.put(targ, num - 1);                                }    }    }    }        return count;    }}