Leetcode_num12_Search Insert Position

标签：leetcode   二叉树   

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.

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

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

Leetcode_num12_Search Insert Position

标签：leetcode   二叉树   

原文地址：http://blog.csdn.net/eliza1130/article/details/39554719