Search a 2D Matrix -- leetcode

标签：leetcode   面试   二分法   二维数组   

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。

Search a 2D Matrix -- leetcode

标签：leetcode   面试   二分法   二维数组   

原文地址：http://blog.csdn.net/elton_xiao/article/details/45008787