LeetCode 题解（149）: Search a 2D Matrix II

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.

For 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.

题解：

从右上角开始找。

C++版：

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.For 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.

Java版：

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

Python版：

class Solution:    # @param {integer[][]} matrix    # @param {integer} target    # @return {boolean}    def searchMatrix(self, matrix, target):        row, col = 0, len(matrix[0]) - 1        while row < len(matrix) and col >= 0:            if matrix[row][col] == target:                return True            elif matrix[row][col] > target:                col -= 1            elif matrix[row][col] < target:                row += 1        return False