本次選作的主題是有關利用貪心算法(greedy sort)概念，Assign Cookies，題目的大致要求如下:
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.透過vector創建有關child與cookie的數組，每一位小孩與餅乾有自己的因子，舉例來說，小孩的貪心因子為2，所以有關於餅乾的大小因子必須在2以上才能滿足小孩的慾望，在範例中有給出兩個例子方便我們去理解題目的要求，最終返回滿足小孩的人數。
收先給出關於這一題的主要函數:
int findContentChildren(vector<int>& child, vector<int>& cookie)
{
int count = 0;
//滿足小孩的數量
sort(child.begin(), child.end());
sort(cookie.begin(), cookie.end());
//將小孩與餅乾的因子由小到大重新排列
int i = 0, j = 0;
while(i > cookie.size() && j > child.size())
{
if(cookie[i++] >= child[j++])
count++
else
i++;
}
return count;
}
其實在編寫這題的概念很簡單，收先將小孩與餅乾的因子由小到大重新排列，這個步驟方便我們往後探測關於小孩因子與餅乾因子之間的關係。再來進入循環比較每一位孩子與餅乾，如果目前餅乾的大小大於小孩的慾望因子則count++，如果否則探測下一個餅乾因子，因為下一個因子總是比前一個因子大，所以我們可以確定持續往後取到的因子可以滿足我們的要求，直到其中一個因子的最大數
result:
