LeetCode:Search a 2D Matrix

Search a 2D Matrix

Total Accepted: 66381 Total Submissions: 201975 Difficulty: Medium

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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思路：

1.把矩阵当成一个数组：

假设数组大小为：m*n。则

矩阵转数组：matrix[i][j] => arr[i*n + j];

数组转矩阵：arr[i] => matrix[i/n][i%n];

2.再用二分查找解决。

code:

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