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455. Assign Cookies ​

题目 ​

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 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

 

Note: This question is the same as 2410: Maximum Matching of Players With Trainers.

Related Topics
  • 贪心
  • 数组
  • 双指针
  • 排序

  • 👍 932
  • 👎 0

    • 思路 ​

    将饼干和贪心数组升序排列，优先满足贪心数小的人

    解法 ​

    python
    class Solution:
    def findContentChildren(self, g: List[int], s: List[int]) -> int:
        sorted_g = sorted(g)
        sorted_s = sorted(s)

        res = 0
        j = 0
        for i, value in enumerate(sorted_g):
            while j < len(sorted_s):
                if sorted_s[j] >= value:
                    res += 1
                    j += 1
                    break
                else:
                    j += 1
        return  res

    复杂度分析 ​

    • 时间复杂度 O(mlonm + nlogn)
    • 空间复杂度 O(m + n)

    解法二 ​

    py
    # leetcode submit region begin(Prohibit modification and deletion)
class Solution:
    def findContentChildren(self, g: List[int], s: List[int]) -> int:
        g.sort()
        s.sort()
        i = j = 0
        while i < len(g) and j < len(s):
            if s[j] >= g[i]:
                i += 1

            j +=1

        return i

        
# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(mlonm + nlogn)
    • 空间复杂度 O(1)