## 2019 年 04 月 11 日 • 阅读: 1011 • LeetCode • 阅读设置

• Easy
• Accepted：60,685
• Submissions：125,878

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.

### 代码

• Runtime: 44 ms, faster than 74.03% of C++ online submissions for Assign Cookies.
• Memory Usage: 10.4 MB, less than 76.00% of C++ online submissions for Assign Cookies.
class Solution {
public:
int findContentChildren(vector<int>& g, vector<int>& s) {
int n = g.size(), m = s.size();
sort(g.begin(), g.end(), greater <>()); // 从大到小排序
sort(s.begin(), s.end(), greater <>());
int ans = 0;
int i = 0, j = 0;
while (i < n && j < m)
{
if (s[j] >= g[i])
{
++i, ++j;
++ans;
}
else
++i;
}
return ans;
}
};
• Runtime: 876 ms, faster than 5.04% of C++ online submissions for Assign Cookies.
• Memory Usage: 10.1 MB, less than 100.00% of C++ online submissions for Assign Cookies.
class Solution {
public:
int findContentChildren(vector<int>& g, vector<int>& s) {
int n = g.size(), m = s.size();
vector<bool>vis(m + 1, false);
sort(g.begin(), g.end());
sort(s.begin(), s.end());
int ans = 0;
for (int i = n - 1; i >= 0; --i)
{
for (int j = m - 1; j >= 0; --j)
{
if (!vis[j] && s[j] >= g[i])
{
++ans;
vis[j] = true;
break;
}
}
}
return ans;
}
};

The end.
2019年4月11日 星期四