题目描述

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.

解法一

思路

这题可以换个说法：对于两个数组A和B，求满足A[i] <= B[j]的(i, j)的最多数目。

这样就是一个典型的贪心问题了。不妨先把A和B都排序，然后从前往后遍历A中元素A[i]，并记录B中当前位置j（初始为0），对于每个元素，从j开始遍历B中元素，当满足A[i] <= B[j]时，记录此组合满足，并将j增长1（B[j]不能重复用）。这个算法的关键点就在于对于A中元素，在B中查找时总是从j开始，而且总是在第一个匹配时停下，可以手动模拟一下。

Python

class Solution:
    def findContentChildren(self, g, s):
        """
        :type g: List[int]
        :type s: List[int]
        :rtype: int
        """
        g = sorted(g)
        s = sorted(s)
        i = 0
        count = 0
        for item in g:
            while i < len(s) and item > s[i]:
                i += 1
            if i == len(s):
                break
            count += 1
            i += 1
        return count

Java

public class Solution {
    public int findContentChildren(int[] g, int[] s) {
        Arrays.sort(g);
        Arrays.sort(s);
        int i = 0;
        int count = 0;
        for (int item : g) {
            while (i < s.length && item > s[i]) {
                i++;
            }
            if (i == s.length) {
                break;
            }
            count++;
            i++;
        }
        return count;
    }
}

C++

class Solution {
public:
    int findContentChildren(vector<int> &g, vector<int> &s) {
        sort(g.begin(), g.end());
        sort(s.begin(), s.end());
        int i = 0;
        int count = 0;
        for (auto item : g) {
            while (i < s.size() && item > s[i]) {
                i++;
            }
            if (i == s.size()) {
                break;
            }
            count++;
            i++;
        }
        return count;
    }
};