[LeetCode] 74. Search a 2D Matrix 解题思路

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.

问题：在给定的已排序的二维矩阵中，判断是否包含某个整数。

思路：看到二维数组，首先想到的是二维数组和 一维数组可以直接转换，arr[i][j] 等于 arr[i*col + j]。而在一个已排序的一维数组中搜索一个元素，可以采用分治(Divide and Conquer)思想，更具体些就是二分搜索(Binary Search) 算法。

将上面的结合起来就是，先实现一维数组的二分搜索算法，然后将算法中的一位转换为二维数组即可。

bool searchMatrix(vector<vector<int>>& matrix, int target) {

    if (matrix.size() == 0) {
        return false;
    }
    
    int col = (int)matrix[0].size();
    int row = (int)matrix.size();
    
    if (col == 1 && row == 1) {
        return ( matrix[0][0] == target );
    }
    
    int lm = 0;
    int rm = col * row - 1;
    
    while (lm < rm ) {
        
        
        if (lm + 1 == rm) {
            int quoL = lm / col;
            int remL = lm % col;
            
            int quoR = rm / col;
            int remR = rm % col;
            
            if (matrix[quoL][remL] == target || matrix[quoR][remR] == target) {
                return true;
            }else{
                return false;
            }
        }
        
        int midm = (lm + rm) / 2;
        
        int quo = midm / col;
        int rem = midm % col;
        
        if (matrix[quo][rem] == target) {
            return true;
        }
        
        if (matrix[quo][rem] < target) {
            lm = midm;
        }else{
            rm = midm;
        }
    }
    
    return false;
}