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。

这里用到的是二段式，二分法查找， 即循环中，只有2个分支，省掉 判等的分支。

在第一次二分退出循环后，需要判断，待查找值，位于当前行前，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;    }};

此题可以把二维数组，当成一个一维有序数组进行折半查找。

但需要一个将一维坐标，转换成二维坐标的过程。 将需要额外进/和%的运算。