LC74. 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.

Example 1:

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

Example 2:

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

代码

双指针法

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int tar) {
        if (matrix.empty() || matrix[0].empty())
            return false;
        int n = matrix.size(), m = matrix[0].size();
        cout << n << m << endl;
        int i = 0, j = m - 1;
        while (i < n && j >= 0) {
            if (tar == matrix[i][j])
                return true;
            if (tar > matrix[i][j])
                i++;
            else
                j--;
        }
        return false;
    }
};

二分法

(二分) $O(logn)O(logn)$
我们可以想象把整个矩阵，按行展开成一个一维数组，那么这个一维数组单调递增，然后直接二分即可。
二分时可以通过整除和取模运算得到二维数组的坐标。

时间复杂度分析：二分的时间复杂度是 $O(logn^2)=O(logn)$。

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int tar) {
        if(matrix.empty()||matrix[0].empty()) return false;
        int n = matrix.size(), m = matrix[0].size();
        int l = 0, r = m - 1, mid;
        while(l < r) {
            mid = (l + r) >> 1;
            if(matrix[mid/m][mid%m] <= tar) r = mid;
            else l = mid + 1;
        }
        return matrix[l/m][l%m] == tar;
    }
};