题目链接: https://leetcode-cn.com/problems/search-insert-position/
题目描述:
给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。
你可以假设数组中无重复元素。
示例:
示例 1:
输入: [1,3,5,6], 5
输出: 2
示例 2:
输入: [1,3,5,6], 2
输出: 1
示例 3:
输入: [1,3,5,6], 7
输出: 4
示例 4:
输入: [1,3,5,6], 0
输出: 0
思路:
二分法
思路1:二分法进行时判断
思路2:二分法执行完毕判断
关于二分搜索,请看这篇[二分搜索](二分查找有几种写法？它们的区别是什么？ - Jason Li的回答 - 知乎
https://www.zhihu.com/question/36132386/answer/530313852),读完之后,可以加深二分搜索的理解!
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代码:
思路一
python
class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
left = 0
right = len(nums) - 1
while left <= right:
mid = left + (right - left) // 2
if nums[mid] == target:return mid
elif nums[mid] < target:
left = mid + 1
else:
right = mid - 1
return left
java
class Solution {
public int searchInsert(int[] nums, int target) {
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) return mid;
else if (nums[mid] > target) right = mid - 1;
else left = mid + 1;
}
return left;
}
}
思路2
python
class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
left = 0
right = len(nums)
while left < right:
mid = left + (right - left) // 2
if nums[mid] < target:
left = mid + 1
else:
right = mid
return left
java
class Solution {
public int searchInsert(int[] nums, int target) {
int left = 0;
int right = nums.length;
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid] < target) left = mid + 1;
else right = mid;
}
return left;
}
}