Binary Search

所有的binary search 问题都可以变换成找 first xxx 或者last xxx的问题。

注意是return first, last 还是 any.

first： 1 end 前移 （mid = target， end=mid)

          2 最后先check start =target

last :  1 start 后移

         2 最后先check end =target

For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity. If the target number does not exist in the array, return -1.

<span style="color:#666e70;">def binarySearch(self, nums, target):        if len(nums) == 0:            return -1        start, end = 0, len(nums)-1        while </span><span style="color:#ff0000;">start + 1 < end:</span><span style="color:#666e70;">            mid = (start + end)/2            if nums[mid] < target:                start = mid            else:                </span><span style="color:#ff0000;">end = mid</span><span style="color:#666e70;"> # end move forward if nums[mid] == target 否则会略过mid之前的target        if nums[start] == target:            return start        if nums[end] == target:            return end        return -1</span>

For a given sorted array (ascending order) and a target number, find the last index of this number in O(log n) time complexity. If the target number does not exist in the array, return -1.

def binarySearchLast(nums, target):        if len(nums) == 0:            return -1        start, end = 0, len(nums) - 1        while start + 1 < end:            mid = (start + end)/2            if nums [mid] > target:                end = mid            else:                start = mid        if nums[end] == target:            return end        if nums[start] == target:            return start        return -1 

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].

其实就是找到first position 和 last position 就可以了

def searchRange(self, A, target):        if len(A) == 0:            return [-1, -1]                start, end = 0, len(A) - 1        while start + 1 < end:            mid = (start + end) / 2            if A[mid] < target:                start = mid            else:                end = mid                if A[start] == target:            leftBound = start        elif A[end] == target:            leftBound = end        else:            return [-1, -1]                start, end = leftBound, len(A) - 1        while start + 1 < end:            mid = (start + end) / 2            if A[mid] <= target:                start = mid            else:                end = mid        if A[end] == target:            rightBound = end        else:            rightBound = start        return [leftBound, rightBound]        

3 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 from left to right.The first integer of each row is greater than the last integer of the previous row.

可以用两次BS, 一次确定row, 一次确定 column.

或者把它连起来看成一个ascending array,然后直接bs,注意比较 A[mid] 和 target时如何把mid 转化为matrix 里的下标

def searchMatrix(self, matrix, target):         # Solution 2 binary search once, treated as an ascending array                if len(matrix) == 0:            return False        m = len(matrix)        n = len(matrix[0])        start, end = 0, m*n - 1        while start + 1 < end:            mid = (start + end) / 2            x = mid / n            y = mid % n            if matrix[x][y] < target:                start = mid            else:                end = mid        x, y = start / n, start % n        if matrix[x][y] == target:            return True        x, y = end / n, end % n        if matrix[x][y] == target:            return True        return False