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 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.sln
通過簡單分析，可以發現這道題用貪婪算法來實現的話，就是對於每個小孩，儘量找比他的g值大的s值列表中最小的s值。
那麼我們首先對g和s兩個列表進行升序排序，分別從頭開始遍歷g和s兩個數組，對於每個g和s值，如果g大於s，則當前s無法滿足當前g，s指向下一個值。如果g小於等於s，那麼當前s是最適合當前g的值（因爲s和g都已經排過序了），因此s和g都指向下一個值。這樣下來，對於排序後的s和g，我們只需要進行一次遍歷就可以找出結果。遍歷的時間複雜度爲O(n)，而排序的複雜度爲O(nlogn)。因此整個算法的複雜度爲O(nlogn)。python實現如下：
class Solution(object):
def findContentChildren(self, g, s):
"""
:type g: List[int]
:type s: List[int]
:rtype: int
"""
if len(g) == 0 or len(s) == 0:
return 0
g = sorted(g)
s = sorted(s)
g_index = 0
s_index = 0
count = 0
while g_index < len(g) and s_index < len(s):
if g[g_index] <= s[s_index]:
g_index += 1
s_index += 1
count += 1
else:
s_index += 1
return count提交上述算法後發現僅僅比75%+的提交快，然而快速瀏覽了下discussion卻發現大家都是在說這種O(nlogn)的實現，所以暫時還沒有進一步優化的思路。