水题,只是思想还是实用的. 当然能够新建出一个数组.比較着拷贝过去.可是没有必要啊亲.想想为什么用源数组会麻烦,由于确定了前面的数.为了后面的数字不被覆盖,要依次往后移,转念一想,先确定后面的数字.就不用操心会覆盖的问题了.像不像杨辉三角形中仅仅要求O(N)空间时的思想,好多好多样例. class Solution { public: void merge(int A[], int m, int B[], int n) { int i=m-1, j = n-1, k = m+n-1; while…
Search in Rotated Sorted Array II Follow up for "LeetCode: Search in Rotated Sorted Array 解题报告":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the arr…
LeetCode:Search in Rotated Sorted Array Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target value to search. If found in the array return its index, other…
LeetCode:Remove Duplicates from Sorted Array Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this in place with constan…
Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory. For example,Given input array A = […
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 Java解法 Python解法 日期 题目地址:https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/#/description 题目描述 Given an array where elements are sorted in ascending order,…