标签：cookies ++ ann nts end turn ber size compare
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 };
标签：cookies ++ ann nts end turn ber size compare
原文地址：http://www.cnblogs.com/amazingzoe/p/6072258.html
