# LeetCode 35：Search Insert Position

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.

Example 1:

`Input: [1,3,5,6], 5 Output: 2`

Example 2:

`Input: [1,3,5,6], 2 Output: 1`

Example 3:

`Input: [1,3,5,6], 7 Output: 4`

Example 4:

`Input: [1,3,5,6], 0 Output: 0`

`package com.lihuia.leetcode;/** * Copyright (C), 2018-2019 * FileName: ThirtyFive * Author: lihui * Date: 2019-08-17 */public class ThirtyFive { public int searchInsert(int[] nums, int target) { int length = nums.length; for (int i = 0; i < length; i++) { if (nums[i] >= target) { return i; } } return length; }}`

`package com.lihuia.leetcode;/** * Copyright (C), 2018-2019 * FileName: ThirtyFive * Author: lihui * Date: 2019-08-17 */public class ThirtyFive { public int searchInsert(int[] nums, int target) { int left = 0; int right = nums.length - 1; //{1,3,5,6} 2 while (left <= right) { int middle = (left + right) / 2; if (nums[middle] == target) { return middle; } else if (nums[middle] > target){ right = middle - 1; } else { left = middle + 1; }  } return left; }}`

Python实现，同样山寨一版本

`#!/usr/bin/env pythonclass Solution(object): def searchInsert(self, nums, target): """ :type nums: List[int] :type target: int :rtype: int """ left = 0 right = len(nums) - 1 while left <= right: middle = (left + right) / 2 if nums[middle] == target: return middle elif nums[middle] > target: right = middle - 1 else: left = middle + 1 return left`

OVER