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  • 74. Search a 2D Matrix

    Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

    • Integers in each row are sorted from left to right.
    • The first integer of each row is greater than the last integer of the previous row.

    For example,

    Consider the following matrix:

    [
      [1,   3,  5,  7],
      [10, 11, 16, 20],
      [23, 30, 34, 50]
    ]
    

    Given target = 3, return true.

    1. 

    转换成一位数组下标,二分查找。

    bool searchMatrix01(vector<vector<int>>& matrix, int target) {
        int row = matrix.size();
        int col = row>0 ? matrix[0].size() : 0;
            
        int len = row * col;
        int low = 0, high = len -1;
        while (low <= high) {
                
            int mid = low + (high - low) / 2;
            int r = mid / col;
            int c = mid % col;
     
            int n = matrix[r][c];
            if (n == target) return true;
            if (n < target) low = mid+1;
            else high = mid -1;
        }
        return false;
    }

    2.

    行和列分别二分查找。

    bool searchMatrix02(vector<vector<int> > &matrix, int target) {
            
        int idx = vertical_binary_search(matrix, target);
        if (idx<0){
            return false;
        }
            
        idx = binary_search(matrix[idx], target);
            
        return (idx < 0 ? false : true);
            
    }
    
    int vertical_binary_search(vector< vector<int> > v, int key){
        int low = 0;
        int high = v.size()-1;
        while(low <= high){
            int mid = low + (high-low)/2;
            if (v[mid][0] == key){
                return mid;
            }
            if (key < v[mid][0]){
                high = mid - 1;
                continue;
            }
            if (key > v[mid][0]){
                low = mid + 1;
                continue;
            }
        }
        return low-1;        
    }
        
    int binary_search(vector<int> v, int key) {
        int low = 0;
        int high = v.size()-1;
        while(low <= high){
            int mid = low + (high-low)/2;
            if (v[mid] == key){
                return mid;
            }
            if (key < v[mid]){
                high = mid - 1;
                continue;
            }
            if (key > v[mid]){
                low = mid + 1;
                continue;
            }
        }
        return -1;
    }
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  • 原文地址:https://www.cnblogs.com/argenbarbie/p/5284305.html
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