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  • 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 };
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  • 原文地址:https://www.cnblogs.com/amazingzoe/p/6072258.html
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