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.

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首先二分查找每行的第一个元素,确定了行号后,再在当前行内进行二分查找

值得一提的是其时间复杂度为 O(log(m) + log(n)) = O(log(m*n))

bool searchMatrix(vector<vector<int>>& matrix, int target) {
    int beg = 0, mid, end = matrix.size()-1;
    while (beg <= end)
    {
        mid = (beg + end) >> 1;
        if (target < matrix[mid][0])
            end = mid - 1;
        else if (target > matrix[mid][0])
            beg = mid + 1;
        else
            return true;
    }
    int row = end;
    if (row < 0)
        return false;
    beg = 0;
    end = matrix[row].size();
    while (beg <= end)
    {
        mid = (beg + end) >> 1;
        if (target < matrix[row][mid])
            end = mid - 1;
        else if (target > matrix[row][mid])
            beg = mid + 1;
        else
            return true;
    }
    return false;
}
原文地址:https://www.cnblogs.com/sdlwlxf/p/5107588.html