Search a 2D Matrix -- leetcode

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

.

基本思路：

1. 先用折半查找，定位到行。

2. 再用折半查找，定位到列。

這裡用到的是二段式，二分法查找， 即循環中，隻有2個分支，省掉 判等的分支。

在第一次二分退出循環後，需要判斷，待查找值，位于目前行前，或者是前一行。

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        if (matrix.empty() || matrix[0].empty())
            return false;
        
        int low = 0, high = matrix.size()-1;
        while (low < high) {
            const int mid = low + (high - low) / 2;
            if (matrix[mid][0] < target)
                low = mid + 1;
            else
                high = mid;
        }

        const int row = low && matrix[low][0] > target ? low - 1: low;
        low = 0, high = matrix[row].size()-1;
        while (low < high) {
            const int mid = low + (high - low) / 2;
            if (matrix[row][mid] < target)
                low = mid + 1;
            else
                high = mid;
        }
        
        return matrix[row][low] == target;
    }
};
           

此題可以把二維數組，當成一個一維有序數組進行折半查找。

但需要一個将一維坐标，轉換成二維坐标的過程。 将需要額外進/和%的運算。