题目: 搜索插入位置 给定一个排序数组和一个目标值,如果在数组中找到目标值则返回索引.如果没有,返回到它将会被按顺序插入的位置. 你可以假设在数组中无重复元素. 样例 [1,3,5,6],5 → 2 [1,3,5,6],2 → 1 [1,3,5,6], 7 → 4 [1,3,5,6],0 → 0 解题: 二分法直接搞,找到了返回下标,找不到结束时候的start ==end就是要插入的下标,非递归程序. Java程序: public class Solution { /** * param num…

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 →…

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:…

给定一个排序数组和一个目标值,如果在数组中找到目标值则返回索引.如果没有,返回到它将会被按顺序插入的位置.你可以假设在数组中无重复元素.案例 1:输入: [1,3,5,6], 5输出: 2案例 2:输入: [1,3,5,6], 2输出: 1案例 3:输入: [1,3,5,6], 7输出: 4案例 4:输入: [1,3,5,6], 0输出: 0详见:https://leetcode.com/problems/search-insert-position/description/ Java实现: c…

这道题是LeetCode里的第35道题. 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 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 二分查找,简单快捷.每一次将被查找区间分为两块,然后再与该区间…

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:…

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 …

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 →…

问题描述:给定一个有序序列,如果找到target,返回下标,如果找不到,返回插入位置. 算法分析:依旧利用二分查找算法. public int searchInsert(int[] nums, int target) { return binarySearch(nums, 0, nums.length - 1, target); } public int binarySearch(int[] nums, int left, int right, int target) { int mid = (…

找出指定target的位置(没有此数时为按顺序应当位置). public class Solution { /** * param A : an integer sorted array * param target : an integer to be inserted * return : an integer */ public int searchInsert(int[] A, int target) { if(A == null) return -1; if(A.length == 0…