【leetcode】Search for a Range

题目描述：

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：

明显的简单二分问题，首先用二分找到一个满足条件的点，然后向两边延展即可

coding=utf-8

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):
        l = len(A)
        left = 0
        right = l-1
        index1 = -1
        index2 = -1
        pos = -1
        while left <= right:
            mid = (left + right) / 2
            if A[mid] == target:
                pos = mid
                break
            elif A[mid] > target:
                right = mid - 1
            else:
                left = mid + 1
        #print pos
        if pos == -1:
            return [-1,-1]
        index1 = index2 = pos
        while A[index1] == target and index1 > 0 and A[index1-1] == target:
            index1 -= 1
        while A[index2] == target and index2 < l-1 and A[index2+1] ==target:
            index2 += 1
        return [index1,index2]

s = Solution()
a = [5, 7, 7, 8, 8, 10]
print s.searchRange(a,8)
a = [1]
print s.searchRange(a,1)