leetcode - 35. Search Insert Position - Easy descrition 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 arr…

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:…

35. 搜索插入位置 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 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 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/prob…

基础题之一,是混迹于各种难题的基础,有时会在小公司的大题见到,但更多的是见于选择题... 题意:在一个有序数列中,要插入数target,找出插入的位置. 楼主在这里更新了<二分查找综述>第一题的解法,比较类似,当然是今天临时写的. 知道了这题就完成了leetcode 4的第二重二分的写法了吧,楼主懒... class Solution { public: int searchInsert(vector<int>& nums, int target) { , r = nums…

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. Here are few examples.[1,3,5,6], 5 → 2[1,3,5,6], 2 →…

题目链接: https://leetcode.com/problems/search-insert-position/?tab=Description   在给定的有序数组中插入一个目标数字,求出插入该数字的下标 由于该数组是已经排好序的数组,可以利用二分查找.   二分查找的返回结果: 1. 当查找的数字在数组中时,返回第一次出现的下标       2. 当查找的数字不存在时,返回 - pos - 1(即 应当插入位置的相反数再减去      参考代码:  package leetcode_5…

转载:https://leetcode.windliang.cc/leetCode-35-Search-Insert-Position.html    思路 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 n…

35. Search Insert Position Description 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: In…

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 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 python的list是有index()方法的,如果目标值在数组中就很简单 如果没有的话就从头线性扫描一遍,当…

35. 搜索插入位置 二分,太简单,没啥好说的 class Solution { public int searchInsert(int[] nums, int target) { if (nums.length == 0) return 0; int i = 0, j = nums.length; int mid = (i + j) / 2; while (i < j) { if (nums[mid] == target) { return mid; } else if (nums[mid]…