LeetCode OJ-74.Search a 2D Matrix

LeetCode OJ-74.Search a 2D Matrix

題目描述

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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題目理解

​ 在一個二維矩陣中，求解target是否存在。可以將二維轉化爲一維，在一維數組中去尋找。時間複雜度爲O(m * n)，空間複雜度爲O(1)。具體代碼如下：

Code

bool search_matrix(vector<vector<int>> &matrix, int target)
{
    int row = (int) matrix.size();
    int col = (row > 0) ? (int) matrix[0].size() : 0;  //判斷是否是空vector
    int sz = row * col;
    int i;
    for (i = 0; i < sz; ++i) {
        if (matrix[i / col][i % col] == target) {  //將二維轉化爲一維
            return true;
        }
    }

    return false;
}