題目
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.
解法思路（一）
思路描述
- 先把餅干數(shù)組 s 和貪心數(shù)組 g 排序；
- 看最大的餅干能不能滿足最貪心的小孩影晓，如果能镰吵，就看次大的餅干能不能滿足次貪心的孩子；
- 如果最大的餅干不能滿足最貪心的孩子挂签，就看能不能滿足次貪心的孩子疤祭；
解法實(shí)現(xiàn)（二）
時間復(fù)雜度:
- O(nlogn)；
空間復(fù)雜度
- O(1)饵婆；
代碼實(shí)現(xiàn)
package leetcode._455;
import java.util.Arrays;
public class Solution455_1 {
public int findContentChildren(int[] g, int[] s) {
Arrays.sort(g);
Arrays.sort(s);
int gi = g.length - 1, si = s.length - 1;
int res = 0;
while(gi >= 0 && si >= 0){
if(s[si] >= g[gi]){
res ++;
si --;
}
gi --;
}
return res;
}
public static void main(String[] args) {
int g1[] = {1, 2, 3};
int s1[] = {1, 1};
System.out.println((new Solution455_1()).findContentChildren(g1, s1));
int g2[] = {1, 2};
int s2[] = {1, 2, 3};
System.out.println((new Solution455_1()).findContentChildren(g2, s2));
}
}
