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.

 

这道题是二分查找Search Insert Position的题目，因为矩阵是行有序并且列有序，查找只需要先按行查找，定位出在哪一行之后再进行列查找即可，所以就是进行两次二分查找。时间复杂度是O(logm+logn)，空间上只需两个辅助变量，因而是O(1)，代码如下：

public boolean searchMatrix(int[][] matrix, int target) {
    if(matrix == null || matrix.length==0 || matrix[0].length==0)
        return false;
    int l = 0;
    int r = matrix.length-1;
    while(l<=r)
    {
        int mid = (l+r)/2;
        if(matrix[mid][0] == target) return true;
        if(matrix[mid][0] > target)
        {
            r = mid-1;
        }
        else
        {
            l = mid+1;
        }
    }
    int row = r;
    if(row<0)
        return false;
    l = 0;
    r = matrix[0].length-1;
    while(l<=r)
    {
        int mid = (l+r)/2;
        if(matrix[row][mid] == target) return true;
        if(matrix[row][mid] > target)
        {
            r = mid-1;
        }
        else
        {
            l = mid+1;
        }
    }
    return false;
}

二分查找是面试中出现频率不低的问题，但是很少直接考二分查找，会考一些变体，除了这道题，还有Search in Rotated Sorted Array和Search for a Range，思路其实差不多，稍微变化一下即可，有兴趣可以练习一下哈。

 

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        if (matrix.empty() || matrix[0].empty()) return false;
        
        const int m = matrix.size(), n = matrix[0].size();
        int start = 0, end = m * n - 1;
        
        while (start <= end) {
            int mid = start + (end - start) / 2;
            int i = mid / n, j = mid % n;
            if (matrix[i][j] == target) return true;
            
            if (matrix[i][j] < target) start = mid + 1;
            else end = mid - 1;
        }
        
        return false;        
    }
};