# DAY 27
0
Software Development

# 原始題目

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

# 解題過程

1. 排序胃口跟餅乾
2. 餅乾從最小到最大，依序餵飽小孩
3. 餵飽後計數器 +1
4. 如果餅乾無法滿足指定的小孩，則跳出迴圈
5. 回傳可滿足的小孩數量
``````class Solution:
def findContentChildren(self, g: List[int], s: List[int]) -> int:
g.sort()
s.sort()

# 可以餵飽的小孩數量
res = 0

# 遍歷所有餅乾
for i in range(len(s)):
# 如果可餵飽的數量小於小孩總數，且餅乾大小大於下一個小孩的胃口。則餵飽數量 +1
if res < len(g) and s[i] >= g[res]:
res += 1
else:
break
return res

``````

# 結果 