[LeetCode] 240. Search a 2D Matrix II

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题：https://leetcode.com/problems/search-a-2d-matrix-ii/description/

题目

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

Consider the following 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]
]
Given target = 5, return true.

Given target = 20, return false.

思路

剑指offer中见到的一道题。从矩阵的 右上角开始探索。若当前 元素比 target小，当前行加一，若大，当前列减一。若 当前元素等于target ，返回True。若 当前 行 或 列 越界，则返回 False。

code

Your runtime beats 79.66 % of python3 submissions.

class Solution:def searchMatrix(self, matrix, target):""":type matrix: List[List[int]]:type target: int:rtype: bool"""n = len(matrix)if n ==0:return Falsem = len(matrix[0])i = 0j = m-1while not( i == n or j == -1 ):if matrix[i][j]>target:j -= 1elif matrix[i][j]<target:i += 1else:return Truereturn False

第二版 java 编写

class Solution {public boolean searchMatrix(int[][] matrix, int target) {int matRow = matrix.length;if(matRow == 0)return false;int matCol = matrix[0].length;int pr = 0;int pc = matCol -1;while(pr<matRow && pc>=0){if(matrix[pr][pc]<target)pr++;else if(matrix[pr][pc]>target)pc--;elsereturn true;}return false;}
}

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