[LeetCode] 35. 搜索插入位置

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;
        
    }
}