LeetCode 240. Search a 2D Matrix II

 

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题目描述：

　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.

思路： 单调的矩阵-->二分筛选

<1> 砍上  二分右列 从上到下找到第一个比target大的

<2> 砍下  二分左列 从下到上找到第一个比target小的

<3> 砍左  二分下行 从左到右找到第一个比target大的

<4> 砍右  二分上行 从右到左找到第一个比target小的

具体代码：

class Solution(object): def searchMatrix(self, matrix, target): """ :type matrix: List[List[int]] :type target: int :rtype: bool """ n = len(matrix) if n==0: return False m = len(matrix[0]) if m==0: return False lr = 0 lc = 0 rr = n-1 rc = m-1
        while lr<=rr and lc<=rc: l = lr r = rr while l<=r: k = (l+r)/2
                if matrix[k][rc] == target: return True elif matrix[k][rc] < target: l = k+1
                else: r = k-1
            if l >= n: return False lr = l l = lr r = rr while l<=r: k = (l+r)/2
                if matrix[k][lc] == target: return True elif matrix[k][lc] < target: l = k+1
                else: r = k-1
            if r <0 : return False rr = r l = lc r = rc while l<=r: k = (l+r)/2
                if matrix[rr][k] == target: return True elif matrix[rr][k] < target: l = k+1
                else : r = k-1
            if l>=m: return False lc = l l = lc r = rc while l<=r: k = (l+r)/2
                if matrix[lr][k] == target: return True elif matrix[lr][k] < target: l = k+1
                else : r = k-1
            if r<0: return False rc = r return False

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