1. 題目描述
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.
2. 樣例
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.
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.
3. 分析
【理解】：給了兩個輸入的數組，分別代表了每個孩子期待分到的cookie的大小和所擁有的cookie的大小情況。因此第一個數組的長度就是孩子的總數，第二個數組的長度就是cookie的總數。我們需要求出怎麼樣分配：才能使最多數量的孩子對cookie的需求得到滿足。
【思路】：我個人覺得用貪心也比較容易想思路：我們將所有的cookie按照大小降序排序，將所有的孩子期待得到cookie的大小也降序排序，接下來兩個數組同時開始遍歷，這樣的做法能夠保證，每一步驟下都可以判斷當前的孩子是否可以滿足心願，因爲當前的cookie也是最大的（比其更大的cookie已經分給了需求更大的孩子）。
如果當前cookie大小大於等於當前孩子的需求：說明這塊cookie是可以滿足條件的，記錄下來；如果當前cookie大小小於當前孩子的需求：那麼說明這個孩子的需求得不到滿足，就只好捨棄他，我們去訪問下一個孩子看看能否滿足他的需求，以求更多的孩子能分到cookie。
4. 源碼
class Solution {
public:
static bool Comp(int a, int b) {
return (a > b);
}
int findContentChildren(vector<int>& g, vector<int>& s) {
sort(g.begin(), g.end(), Comp);
sort(s.begin(), s.end(), Comp);
int counter = 0;
int i = 0, j = 0;
while(i < (int)s.size() && j < (int)g.size()) {
if (s[i] >= g[j]) {
counter++;
i++;
j++;
}
else {
j++;
}
}
return counter;
}
};5. 心得
這道題屬於典型的貪心算法的最簡單應用，之所以簡單是每個孩子最多隻允許被分配一個cookie，如果cookie可以累加，那麼問題就複雜了很多。