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[leetcode]Search for a Range @ Python

时间:2014-06-08 21:03:21      阅读:297      评论:0      收藏:0      [点我收藏+]

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原题地址:https://oj.leetcode.com/problems/search-for-a-range/

题意:

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm‘s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

解题思路:又是二分查找的变形。因为题目要求的时间复杂度是O(log n)。在二分查找到元素时,需要向前和向后遍历来找到target元素的起点和终点。

代码:

bubuko.com,布布扣
class Solution:
    # @param A, a list of integers
    # @param target, an integer to be searched
    # @return a list of length 2, [index1, index2]
    def searchRange(self, A, target):
        left = 0; right = len(A) - 1
        while left <= right:
            mid = (left + right) / 2
            if A[mid] > target:
                right = mid - 1
            elif A[mid] < target:
                left = mid + 1
            else:
                list = [0, 0]
                if A[left] == target: list[0] = left
                if A[right] == target: list[1] = right
                for i in range(mid, right+1):
                    if A[i] != target: list[1] = i - 1; break
                for i in range(mid, left-1, -1):
                    if A[i] != target: list[0] = i + 1; break
                return list
        return [-1, -1]
bubuko.com,布布扣

 

 

[leetcode]Search for a Range @ Python,布布扣,bubuko.com

[leetcode]Search for a Range @ Python

标签:c   style   class   blog   code   java   

原文地址:http://www.cnblogs.com/zuoyuan/p/3775904.html

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