LeetCode (Search a 2D Matrix)

胡为乎遑遑于何之

于 2017-05-13 20:34:14 发布

阅读量236

点赞数

CC 4.0 BY-SA版权

分类专栏： C++

本文链接：https://blog.youkuaiyun.com/m0_37705274/article/details/71893957

C++ 专栏收录该内容

83 篇文章

订阅专栏

本文介绍了一种高效的矩阵搜索算法，该算法能在排序的二维矩阵中快速查找目标值。矩阵的每一行从左到右递增排序，并且每行的第一个元素大于前一行的最后一个元素。文章通过一个示例矩阵和目标值3，演示了如何使用该算法实现查找。

Problem:

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.

Solution:

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