Assign Cookies

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.


Solution1 : brute force, time limited. 

 1 class Solution {
 2 public:
 3     // for every child, find a size s_j which is equal or larger than g_i, and s_j is the minimum among all larger than g_i
 4     int findContentChildren(vector<int>& g, vector<int>& s) {
 5         int contentNumber = 0;
 6         if (g.empty() || s.empty()) return contentNumber;
 7         
 8         for (int i = 0; i < g.size(); i++) {
 9             int minimumSatisfySize = INT_MAX;
10             int minimumSatisfyIndex = 0;
11             for (int j = 0; j < s.size(); j++) {
12                 if (s[j] < minimumSatisfySize && s[j] >= g[i]) {
13                     minimumSatisfySize = s[j];
14                     minimumSatisfyIndex = j;
15                 }
16             }
17             if (minimumSatisfySize != INT_MAX) {
18                 contentNumber++;
19                 s[minimumSatisfyIndex] = INT_MAX;
20             }
21         }
22         return contentNumber;
23     }
24 };

 

Solution 2: first sort two arrays, then use two pointers to compare the elements in each array. 

 1 class Solution {
 2 public:
 3     int findContentChildren(vector<int>& g, vector<int>& s) {
 4         int result = 0;
 5         if (g.empty() || s.empty()) return result;
 6         
 7         sort(g.begin(), g.end());
 8         sort(s.begin(), s.end());
 9         
10         int i = 0, j = 0;
11         int m = g.size(), n = s.size();
12         while (i < m && j < n) {
13             if (g[i] <= s[j]) {
14                 i++;
15                 j++;
16                 result++;
17             } else {
18                 j++;
19             }
20         }
21         return result;
22     }
23 };

 

转载于:https://www.cnblogs.com/amazingzoe/p/6072258.html