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