Leetcode_num12_Search Insert Position

题目：

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

比较直观的方法如下：

class Solution:    # @param A, a list of integers    # @param target, an integer to be inserted    # @return integer    def searchInsert(self, A, target):        rs=0 #结果        L=len(A)        for i in range(L):            if target<A[0]:                rs=0                break            elif target>A[L-1]:                rs=L                break            elif target==A[i]:                rs=i                break            elif target>A[i] and target<A[i+1]:                rs=i+1                break        return rs

但是该方法的复杂度为o(n)

复杂度为o(logn)的方法需要利用二分法

代码如下：

class Solution:    # @param A, a list of integers    # @param target, an integer to be inserted    # @return integer    def searchInsert(self, A, target):        L=len(A)        low=0        high=L-1        while(low<=high):            mid=low+(high-low)/2            if target<A[mid]:                high=mid-1            elif target>A[mid]:                low=mid+1            else:                return mid        return low