Topic 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.
Function to be done:
int findContentChildren(vector<int>& g, vector<int>& s)
illustrate:
1. It is not difficult to understand the meaning of this question. Each child has a required size number, which is an integer, such as 1, 2, and 3. There are a few biscuits, the size is an integer, like 1, 2, 3. Can't break the biscuits.
The output is required to be able to meet the needs of several children.
2. First, sort the required size g, sort the biscuit size s, and then start the comparison from the first position of g and s until g or s is processed.
The code is also very simple, as follows:
int findContentChildren(vector<int>& g, vector<int>& s)
{
sort(g.begin(),g.end());
sort(s.begin(),s.end());
int i= 0 ,j= 0 ;
while (i<g.size()&&j<s.size()) // After processing the needs of all children or there are no more cookies to meet the needs
{
if (s[j]> = g[i])
i ++; // If satisfied, process next i
j++; // If not satisfied, add j all the time
}
return i;
}
The above code measured 40ms, beats 97.41% of cpp submissions.