[LeetCode] 240. Search a 2D Matrix II

轉自LeetCode

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

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
• -109 <= matrix[i][j] <= 109
• All the integers in each row are sorted in ascending order.
• All the integers in each column are sorted in ascending order.
• -109 <= target <= 109

Solution

74. Search a 2D Matrix 的衍生題

差別在於，沒有保證下一行的數字都比上一行大

所以對row和col做binary search 的方法會不行

但單純就row和col本身來看，數值都還是sorted的

所以右上角往左往下，都會是遞減

或者左下角往上往右，都會是遞增

可以從這兩個地方出發來找值，這次選用右上角

如果 target 比目前的值大，到下移動

如果 target 比目前的小，往左移動

kotlin

	class Solution {
	fun searchMatrix(matrix: Array<IntArray>, target: Int): Boolean {
	var row = 0
	var col = matrix[0].lastIndex
	while(row < matrix.size && col > -1) {
	when {
	matrix[row][col] == target -> return true
	matrix[row][col] < target -> ++row
	else -> --col
	}
	}
	return false
	}
	}

view raw search_a_2_matrix_II.kt hosted with ❤ by GitHub