74 Search a 2D Matrix

74. Search a 2D Matrix​
1. Question
Write an efficient algorithm that searches for a value in anmxnmatrix. 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:
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[
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  [1,   3,  5,  7],
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  [10, 11, 16, 20],
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  [23, 30, 34, 50]
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]
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Given target=3, returntrue.
2. Implementation
(1) Binary Search
思路: 因为每一行的第一个数肯定比上一行的最后一个数大，所以matrix[0][0]到matrix[m - 1][n - 1]是严格递增的, 可以直接把matrix降成一维数组，对这个一维数组进行二分查找
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class Solution {
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    public boolean searchMatrix(int[][] matrix, int target) {
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        if (matrix == null || matrix.length == 0) {
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            return false;
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        }
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​
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        int m = matrix.length, n = matrix[0].length;
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        int start = 0, end = m * n - 1, mid = 0;
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​
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        while (start <= end) {
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            mid = start + (end - start) / 2;
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            int row = mid / n;
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            int col = mid % n;
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​
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            if (matrix[row][col] == target) {
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                return true;
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            }
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            else if (matrix[row][col] < target) {
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                start = mid + 1;
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            }
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            else {
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                end = mid - 1;
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            }
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        }
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        return false;
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    }
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}
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3. Time & Space Complexity
时间复杂度O(logmn), m为matrix的行数, row为matrix的列数，空间复杂度O(1)

69 Sqrt(x)

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Contents

74. Search a 2D Matrix

1. Question

2. Implementation

3. Time & Space Complexity

74. Search a 2D Matrix​

1. Question

2. Implementation

3. Time & Space Complexity

74. Search a 2D Matrix