LeetCode #455 Assign Cookies 分发饼干

455 Assign Cookies 分发饼干

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 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:

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.

题目描述:
假设你是一位很棒的家长，想要给你的孩子们一些小饼干。但是，每个孩子最多只能给一块饼干。对每个孩子 i ，都有一个胃口值 gi ，这是能让孩子们满足胃口的饼干的最小尺寸；并且每块饼干 j ，都有一个尺寸 sj 。如果 sj >= gi ，我们可以将这个饼干 j 分配给孩子 i ，这个孩子会得到满足。你的目标是尽可能满足越多数量的孩子，并输出这个最大数值。

注意：

你可以假设胃口值为正。
一个小朋友最多只能拥有一块饼干。

示例:

示例 1:

输入: [1,2,3], [1,1]

输出: 1

解释:
你有三个孩子和两块小饼干，3个孩子的胃口值分别是：1,2,3。
虽然你有两块小饼干，由于他们的尺寸都是1，你只能让胃口值是1的孩子满足。
所以你应该输出1。

示例 2:

输入: [1,2], [1,2,3]

输出: 2

解释:
你有两个孩子和三块小饼干，2个孩子的胃口值分别是1,2。
你拥有的饼干数量和尺寸都足以让所有孩子满足。
所以你应该输出2.

思路:

由于每个小孩最多获得一块饼干, 所以用最小的饼干满足要求最小的人能满足更多的人
先对数组进行排序, 然后用贪心算法分发饼干
时间复杂度O(nlgn), 空间复杂度O(1)

代码:
C++:

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

Java:

class Solution {
    public int findContentChildren(int[] g, int[] s) {
        Arrays.sort(g);
        Arrays.sort(s);
        int result = 0;
        for (int bis : s) {
            if (bis >= g[result]) result++;
            if (result == g.length) return result;
        }
        return result;
    }
}

Python:

class Solution:
    def findContentChildren(self, g: List[int], s: List[int]) -> int:
        g.sort()
        s.sort()
        result = 0
        for bis in s:
            if g[result] <= bis:
                result += 1
            if result == len(g):
                return result
        return result