Task

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.

Example

Example 1:

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Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Example 2:

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Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false

Solution

  • 整个矩阵拉成一维数组是升序排列，故用二分搜索
  • 时间复杂度：O(log(m*n))
  • 空间复杂度：O(1)
  • 实现
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// Runtime: 20 ms, faster than 24.78% of C++ online submissions for Search a 2D Matrix.
// Memory Usage: 11.5 MB, less than 18.00% of C++ online submissions for Search a 2D Matrix.
class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        //边界
        if(matrix.size()==0)
            return false;
        //得到矩阵宽高
        int H=matrix.size(),W=matrix[0].size();
        int l=0,r=H*W;
        while(l<r){
            int m=l+(r-l)/2;
            //将拉成一维矩阵后的位置转换为矩阵中的坐标
            int val=matrix[m/W][m%W];
            if(val<target)
                l=m+1;
            else if(val>target)
                r=m;
            else
                return true;
        }
        return false;
    }
};