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.
已知一些孩子和一些糖果,每个孩子有需求因子g,每个糖果有大小s,当某些糖果的大小s>=某个孩子的需求因子g时,代表该糖果可以满足孩子;求使用这些糖果,最多能满足多少孩子?(注意:某个孩子最多只能有1个糖果满足)
解法:先对g, s排序,然后使用贪心法。 某个糖果如果不能满足某个孩子,则该糖果也一定不能满足需求因子更大的孩子。某个孩子可以用更小的糖果满足,则没必要更大糖果满足,因为可以保留更大的糖果满足需求因子更大的孩子(贪心)。
class Solution {
public:
int findContentChildren(vector<int>& g, vector<int>& s) {
sort(g.begin(),g.end());
sort(s.begin(), s.end());
int child = 0, cookie = 0;
while (child < g.size() && cookie < s.size())
{
if (g[child] <= s[cookie])
{
child++;
}
cookie++;
}
return child;
}
};