74/240-搜索二维矩阵

LeetCode

发布日期: 2021-06-03

LeetCode 74. Search a 2D Matrix

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.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

method: 二分

注意二维下标的查找方式
通过一维下标除以列数、模以列数索引二维矩阵

0   1   2   3
4   5   6   7
8   9  10  11

一维下标9对应的二维下标是：[9 / 4][9 % 4] = [2][1]

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

时间复杂度：$O(log(mn))$
空间复杂度：$O(1)$

LeetCode 240. Search a 2D Matrix II

LeetCode-240

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

method: 单调性扫描（Z字形查找）

从右上角看，左边的数都比它小，下边的数都比它大
如果大于target，就往左移
如果小于target，就往下移

时间复杂度：$O(n + m)$

bool searchMatrix(vector<vector<int>>& matrix, int target) {
    int m = matrix.size(), n = matrix[0].size();
    int l = 0, r = n - 1;
    while (l < m && r >= 0) {
        if (matrix[l][r] < target) l++;         // 小了就往下移
        else if (matrix[l][r] > target) r--;    // 大了就往左移
        else return true;
    }
    return false;
}