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中文文档

Description

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 g[i], which is the minimum size of a cookie that the child will be content with; and each cookie j has a size s[j]. If s[j] >= g[i], 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.

 

Example 1:

Input: g = [1,2,3], s = [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: g = [1,2], s = [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.

 

Constraints:

  • 1 <= g.length <= 3 * 104
  • 0 <= s.length <= 3 * 104
  • 1 <= g[i], s[j] <= 231 - 1

Solutions

Python3

class Solution:
    def findContentChildren(self, g: List[int], s: List[int]) -> int:
        g.sort()
        s.sort()
        j = 0
        for i, v in enumerate(g):
            while j < len(s) and s[j] < v:
                j += 1
            if j >= len(s):
                return i
            j += 1
        return len(g)

Java

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

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, j = 0;
        for (; i < g.size(); ++i) {
            while (j < s.size() && s[j] < g[i]) {
                ++j;
            }
            if (j >= s.size()) {
                break;
            }
            ++j;
        }
        return i;
    }
};

Go

func findContentChildren(g []int, s []int) int {
	sort.Ints(g)
	sort.Ints(s)
	i, j := 0, 0
	for ; i < len(g); i++ {
		for ; j < len(s) && s[j] < g[i]; j++ {
		}
		if j >= len(s) {
			break
		}
		j++
	}
	return i
}

JavaScript

/**
 * @param {number[]} g
 * @param {number[]} s
 * @return {number}
 */
var findContentChildren = function (g, s) {
    g.sort((a, b) => a - b);
    s.sort((a, b) => a - b);
    let i = 0;
    let j = 0;
    for (; i < g.length; ++i) {
        while (j < s.length && s[j] < g[i]) {
            ++j;
        }
        if (j >= s.length) {
            break;
        }
        ++j;
    }
    return i;
};

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