Leetcode#34 Search for a Range

 

題目描述：

 

Given an array of integers nums sorted in ascending order, 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].

 

 

分析：

 

Search for a range:

pre: a inc list of int and the target val
post: the range for the val in the list of [-1,-1]

與普通的二分法要進行改進：

 

 

    當mid的值與val值相同時，要從head->mid和從mid->tail中選出新的前後值

這裏採取先選擇head值，重複操作，直到head和tail對應的數據等於目標值val，退出循環。

 

 

 

 

class Solution:
    def searchRange(self, nums, val):
        head = 0
        tail = len(nums)-1
        mid = (head+tail)//2
        while head < tail:
            flag = 0 
            if ( nums[mid] < val ):
                head = mid+1
                mid = (head+tail)//2
            elif nums[mid] > val :
                tail = mid-1
                mid = (head+tail)//2
            elif nums[head] < val :
                mid = (head+mid)//2
            elif nums[tail] > val :
                mid = (tail+mid)//2+1
            else :
                break 
        ans = []
        if ( head >= len(nums) or nums[head] != val ):
            ans.append(-1)
            ans.append(-1)
        else :
            ans.append( head)
            ans.append( tail )
        return ans