[LeetCode] 34. Find First and Last Position of Element in Sorted Array == [LintCode] 61. Search for a Range_Easy tag: Binary Search

Description

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

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

Example

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

Challenge

O(log n) time.

这个题目的思路就是用两个binary search, 分别求first index 和last index.

Code   其实可以利用helper funciton来去优化code.

class Solution:
    def searchRange(self, A, target):
        # write your code here
        if not A: return [-1,-1]
        l, r , ans = 0, len(A) -1, [-1,-1]
        while l + 1 < r:
            mid = l + (r - l)//2
            if A[mid] < target:
                l = mid
            elif A[mid] > target:
                r = mid
            else:
                r = mid
        if A[l] == target:
            ans[0] = l
        elif A[r] == target:
            ans[0] = r
        else:
            return ans

        # find last index
        l, r = 0, len(A) -1
        while l + 1 < r:
            mid = l + (r - l)//2
            if A[mid] < target:
                l = mid
            elif A[mid] > target:
                r = mid
            else:
                l = mid
        if A[r] == target:
            ans[1] = r
        elif A[l] == target:
            ans[1] = l
        else:
            return ans
        return ans

去掉不必要的行

class Solution:
    def searchRange(self, A, target):
        # write your code here
        if not A: return [-1,-1]
        l, r , ans = 0, len(A) -1, [-1,-1]
        while l + 1 < r:
            mid = l + (r - l)//2
            if A[mid] < target:
                l = mid
            else:
                r = mid
        if A[l] == target: ans[0] = l
        elif A[r] == target: ans[0] = r
        else:
            return ans

        # find last index
        l, r = 0, len(A) -1
        while l + 1 < r:
            mid = l + (r - l)//2
            if A[mid] <= target:
                l = mid
            else:
                r = mid
        if A[r] == target: ans[1] = r
        elif A[l] == target: ans[1] = l
        return ans

Use helper function to make the code even shorter.

class Solution:
    def searchRange(self, A, target):
        l, r = 0, len(A) - 1
        if not A or nums[l] > target or nums[r] < target:
            return [-1, -1]
        def helper(points, pos, target):
            while points[0] + 1 < points[1]:
                mid = points[0] + (points[1] - points[0])//2
                if A[mid] > target:
                    points[1] = mid
                elif A[mid] < target:
                    points[0] = mid
                else:
                    points[pos] = mid
            if A[points[1 - pos]] == target: return points[1 - pos]
            if A[points[pos]] == target: return points[pos]
            return -1
        return [helper([l, r], 1, target), helper([l, r], 0, target)]