标签：kth amp ted ret etc while self color 第一题
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3, return true.
第一题的条件比第二题还强，也就是本行的第一个元素比上一行的最后一个元素要大。所以如果第一题把2D矩阵拉平成一个1D list，这个list是一个sorted list。可以再用二分法查找。
1 class Solution(object): 2 def searchMatrix(self, matrix, target): 3 """ 4 :type matrix: List[List[int]] 5 :type target: int 6 :rtype: bool 7 """ 8 if not matrix: 9 return False 10 11 nums = [] 12 for elem in matrix: 13 nums += elem 14 n = len(nums) 15 16 low, high = 0, n-1 17 while low <= high: 18 mid = (low+high)//2 19 if target < nums[mid]: 20 high = mid -1 21 elif target > nums[mid]: 22 low = mid + 1 23 else: 24 return True 25 return False
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
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.
第二题不能像第一题那样把2D matrix拉成一个1D list。可以使用类似 kth smallest element in a sorted matrix 中的类似查找方法。
1 def searchMatrix(self, matrix, target): 2 """ 3 :type matrix: List[List[int]] 4 :type target: int 5 :rtype: bool 6 """ 7 m, n = len(matrix), len(matrix[0]) 8 i, j = m - 1, 0 9 10 while i >= 0 and j < n: 11 if matrix[i][j] > target: 12 i -= 1 13 elif matrix[i][j] < target: 14 j += 1 15 else: 16 return True 17 return False
Leetcode 74 and 240. Search a 2D matrix I and II
标签：kth amp ted ret etc while self color 第一题
原文地址：http://www.cnblogs.com/lettuan/p/6213470.html
