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LeetCode | 0455. Assign Cookies分发饼干【Python】

LeetCode 0455. Assign Cookies分发饼干【Easy】【Python】【贪心】

Problem

LeetCode

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.
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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.
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问题

力扣

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 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。
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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.
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思路

贪心

先将 贪婪指数饼干大小 从小到大排完序,然后 i, j 指针分别指向 贪婪指数饼干大小 的末尾,如果 贪婪指数 满足小于等于饼干大小,说明此 贪婪指数 可以被满足,于是,计数加 1,j 指针往左移一位。不管 贪婪指数 有没有满足,i 指针都要往左移,因为如果满足,那么可以判断下一位,如果不满足,那么自然也不在考虑之中。

时间复杂度: O(max(len(g), len(s))) 空间复杂度: O(1)

Python代码

class Solution(object):
    def findContentChildren(self, g, s):
        """
        :type g: List[int]
        :type s: List[int]
        :rtype: int
        """
        # 贪心
        cnt = 0
        g, s = sorted(g), sorted(s)  # 从小到大排序
        i, j = len(g)-1, len(s)-1  # 指向末尾
        while min(i, j) >= 0:
            if g[i] <= s[j]:  # 贪婪指数要 <= 饼干大小
                cnt += 1
                j -= 1
            i -= 1  # 不管满不满足 size,i 都要往左移一位
        return cnt
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代码地址

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