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.输入两个正整数数组g和s,使得满足s[i]>=g[j]的pair数最多,求这个pair数。
思路:先将两个数组降序排序,为g和s分别设置游标,当满足s[i]>=g[j]时移动两个游标并计数,否则移动g游标。
int findContentChildren(vector<int>& g, vector<int>& s) {
sort(g.begin(),g.end(),greater<int>());
sort(s.begin(),s.end(),greater<int>());
int i=0,j=0,count=0;
while(i<s.size()&&j<g.size())
{
if(s[i]<g[j])
{
j++;
}
else
{
i++;
j++;
count++;
}
}
return count;
}

本文探讨了如何通过合理分配不同大小的饼干来满足多个孩子的愿望,并最大化满足条件的孩子数量。提出了一个有效的算法解决方案,通过降序排序和双指针技巧来解决这个问题。
1150

被折叠的 条评论
为什么被折叠?



