35. Search Insert Position 搜索插入位置

@TOC

# 题目描述

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
``````

# 解题方法

# 二分查找

``````class Solution(object):
def searchInsert(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
return bisect.bisect_left(nums, target)
``````

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

Java代码如下：

``````public class Solution {
public int searchInsert(int[] nums, int target) {
int left = 0;
int right = nums.length - 1;
int mid = 0;
while(left <= right){
mid = (left + right) / 2;
if(nums[mid] == target){
return mid;
}else if(nums[mid] > target){
right--;
}else{
left++;
}
}
return left;
}
}
``````

# 日期

2017 年 4 月 25 日 2018 年 11 月 21 日 —— 又是一个美好的开始