LeetCode OJ-74.Search a 2D Matrix

题目描述

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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题目理解

​ 在一个二维矩阵中，求解target是否存在。可以将二维转化为一维，在一维数组中去寻找。时间复杂度为O(m * n)，空间复杂度为O(1)。具体代码如下：

Code

bool search_matrix(vector<vector<int>> &matrix, int target){    int row = (int) matrix.size();    int col = (row > 0) ? (int) matrix[0].size() : 0;  //判断是否是空vector    int sz = row * col;    int i;    for (i = 0; i < sz; ++i) {        if (matrix[i / col][i % col] == target) {  //将二维转化为一维            return true;        }    }    return false;}