[LeetCode] 240. Search a 2D Matrix Ⅱ（矩阵搜索之二）

• Difficulty: Medium

• Related Topics: Binary Search, Divide and Conquer

• Link: https://leetcode.com/problems/search-a-2d-matrix-ii/

Description

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:
设计一个高效算法，判断一个给定值 target 是否存在于给定的 m x n 整数矩阵 matrix 中。matrix 具有以下属性：

• Integers in each row are sorted in ascending from left to right.
  每一行上的整数从左到右升序排序。
• Integers in each column are sorted in ascending from top to bottom.
  每一列上的整数从上到下升序排序。

Examples

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

Example 2

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 = 20
Output: false

Constraints

• m == matrix.length
• n == matrix[i].length
• 1 <= n, m <= 300
• -1e9 <= matix[i][j] <= 1e9
• All the integers in each row are sorted in ascending order.
• All the integers in each column are sorted in ascending order.
• -1e9 <= target <= 1e9

Solution

这题的关键点在于初始点的选取，一开始我傻乎乎地把起始点放在正中心，结果发现怎么搜也搜不了。discussion 里给出的一个解法是以左上角作为起始点：

• 若目标值比当前值大，说明目标值不在当前值所在的这一行

• 若目标值比当前值小，说明目标值不在当前值所在的这一列

代码如下：

class Solution {
    fun searchMatrix(matrix: Array<IntArray>, target: Int): Boolean {
        if (matrix.isEmpty() || matrix[0].isEmpty()) {
            return false
        }
        var row = 0
        var col = matrix[0].lastIndex
        
        while (row < matrix.size && col >= 0) {
            val curValue = matrix[row][col]
            if (target == curValue) {
                return true
            } else if (target > curValue) {
                row++
            } else {
                col--
            }
        }
        return false
    }
}