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.
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 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.
解法1:（C++实现）
思路是：1）先排序
2）最小的饼干给最小满足的孩子，同理后面剩下的饼干就能满足更大满足度的孩子，从而实现贪心，每个人尽可能满足下，能满足更多的孩子
class Solution {
public:
int findContentChildren(vector<int>& g, vector<int>& s)
{
sort(g.begin(),g.end());//g为胃口值
sort(s.begin(),s.end());//s为饼干尺寸
int p = 0;
int ans = 0;
for (int i = 0; i < g.size(); i++)
{
while (p < s.size() && s[p] < g[i])
p++;
if (p == s.size()) break;//跳出循环说明没匹配，进入下一个孩子的胃口值匹配
p++;
ans++;
}
return ans;
}
};
思路是一样的：
class Solution {
public:
int findContentChildren(vector<int>& g, vector<int>& s)
{
sort(g.begin(), g.end());
sort(s.begin(), s.end());
int child = 0;
int cookie = 0;
while( child < g.size() && cookie < s.size())
{
if(s[cookie] >= g[child])
{
child++;
}
cookie++;
}
return child;
}
};
