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  • 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;
    }
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  • 原文地址:https://www.cnblogs.com/sdlwlxf/p/5107588.html
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