import java.util.Arrays; /** * Source : https://oj.leetcode.com/problems/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 a…

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 = …

/** * 无额外空间,只要前n个是不重复的就行,不需要修改后面的数字 * @param nums 已排序的数组 * @return 去除重复数字后的长度 */ public int removeDuplicates(int[] nums) { if(nums == null || nums.length == 0) { return 0; } else if(nums.length == 1) { return 1; } // 使用两个指针 int j = 0; for(int i = 1;…

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 = […

利用堆栈:http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/http://oj.leetcode.com/problems/longest-valid-parentheses/ (也可以用一维数组,贪心)http://oj.leetcode.com/problems/valid-parentheses/http://oj.leetcode.com/problems/largest-rectangle-in-histo…

问题: 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…

463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…

leetcode代码 利用堆栈:http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/http://oj.leetcode.com/problems/longest-valid-parentheses/ (也可以用一维数组,贪心)http://oj.leetcode.com/problems/valid-parentheses/http://oj.leetcode.com/problems/largest-rectang…

利用堆栈:http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/http://oj.leetcode.com/problems/longest-valid-parentheses/ (也可以用一维数组,贪心)http://oj.leetcode.com/problems/valid-parentheses/http://oj.leetcode.com/problems/largest-rectangle-in-histo…

原链接:http://blog.csdn.net/yangliuy/article/details/44514495 注:此分类仅供大概参考,没有精雕细琢.有不同意见欢迎评论~ 利用堆栈:http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/http://oj.leetcode.com/problems/longest-valid-parentheses/ (也可以用一维数组,贪心)http://oj.leetcode.…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 26: Remove Duplicates from Sorted Arrayhttps://oj.leetcode.com/problems/remove-duplicates-from-sorted-array/ Given a sorted array, remove the duplicates in place such that each element appe…

题目来源: https://leetcode.com/problems/remove-duplicates-from-sorted-array/ 题意分析: 给定一个排好序的数组,去除重复的数,返回新数组的长度,不能申请额外的空间,超过新数组长度部分是什么数都无所谓. 题目思路: 这是一个很简单的题目,由于给定的数组已经排序,那么用i,j两个下标,i记录新数组的下标,j是原来数组下标,如果nums[j] != nums[j - 1],那么nums[i] = nums[j],i 和j 都+ 1.最…

/*@Copyright:LintCode@Author:   Monster__li@Problem:  http://www.lintcode.com/problem/remove-duplicates-from-sorted-array@Language: Java@Datetime: 17-03-02 11:26*/ public class Solution {    /**     * @param A: a array of integers     * @return : ret…

LeetCode 题目总结/分类 利用堆栈: http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/ http://oj.leetcode.com/problems/longest-valid-parentheses/ (也可以用一维数组,贪心) http://oj.leetcode.com/problems/valid-parentheses/ http://oj.leetcode.com/problems/large…

介绍    本篇介绍的是标记元素的使用,很多需要找到正确元素都可以将正确元素应该插入的位置单独放置一个标记来记录,这样可以达到原地排序的效果. Start 27.RemoveElement 删除指定元素 描述 Given an array and a value, remove all instances of that value in place and return the new length. Do not allocate extra space for another array,…

题目链接:https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array/ 题目描述: 给定一个排序数组,你需要在原地删除重复出现的元素,使得每个元素只出现一次,返回移除后数组的新长度. 不要使用额外的数组空间,你必须在原地修改输入数组并在使用 O(1) 额外空间的条件下完成. 示例: 示例 1: 给定数组 nums = [1,1,2], 函数应该返回新的长度 2, 并且原数组 nums 的前两个元素被修改为 1, 2.…

一.模板以及题目分类 1.头尾指针向中间逼近 ; ; while (pos1<pos2) { //判断条件 //pos更改条件 if (nums[pos1]<nums[pos2]) pos1++; else pos2--; } 经典的找和的问题都可以从这种思路下手,2数之和,3数之和,还注意要区分是寻找值还是索引(寻找索引则不能排序),是否允许有重复,不允许重复时要怎样避开重复值. 避开重复值的方法,当然,在3sum和4sum中的ij要稍微做修改 && nums[i] == n…

Algorithm 本周的算法题是删除已排序数据中的重复数字(https://leetcode.com/problems/remove-duplicates-from-sorted-array/).这道题比较简单,基本思路是从开头比较,遇到不同的数字,就进行替换 public static int removeDuplicates(int[] nums) { if (nums == null || nums.length == 0) { return 0; } int i = 0; for (i…

1.这个题不难,关键在于把题目意思理解好了.这个题问的不清楚.要求return new length,很容易晕掉.其实就是return 有多少个单独的数. import java.util.Arrays; /* * Question: Given a sorted array,remove the duplicates in place such that each element * appear only once and return the new length * * Do not a…

基础知识 二分非递归写法: int binary_search(const int a[], const int size, const int val) { int lower = 0; int upper = size-1; /* invariant: if a[i]==val for any i, then lower <= i <= upper */ while (lower <= upper) { int i = lower + (upper-lower)>>1;…

https://leetcode.com/problems/remove-duplicates-from-sorted-array/ 双指针,注意初始时左右指针指向首元素! class Solution { public: int removeDuplicates(vector<int>& nums) { ; int i = nums.size(); ) ; ],*r = &nums[]; ]) { while(*l == *r) { r++; } l++; *l = *r;…

  1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = […

problem RemoveDuplicatesfromSortedArray 注意数组为空的情况要首先考虑,并给出返回值: 注意也要同时给出新的数组的数值: 注意数组最后两个元素的处理: class Solution { public: int removeDuplicates(vector<int>& nums) { ) ; ; ; i<nums.size()-; i++) { ]) { nums[len] = nums[i]; len++; } } nums[len] =…

题目链接: https://leetcode.com/problems/remove-duplicates-from-sorted-array/?tab=Description   从有序数组中移除重复数字,并且返回不重复数字的个数   遍历操作: 可以使用新的for循环 for (int n : nums){}   每次进行对比,并且更新第一个遇到不相等的元素的下标为i 对数组进行重新赋值操作   当数组长度大于1时,ans初值为1,当数组长度为0时,返回0   参考代码 :   packag…

引自:http://www.douban.com/note/330562764/ 注:此分类仅供大概参考,没有精雕细琢.有不同意见欢迎评论~ 利用堆栈:http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/http://oj.leetcode.com/problems/longest-valid-parentheses/ (也可以用一维数组,贪心)http://oj.leetcode.com/problems/valid…

binary search 14.https://www.lintcode.com/problem/first-position-of-target/description 74.https://www.lintcode.com/problem/first-bad-version/description Tree 97.https://www.lintcode.com/problem/maximum-depth-of-binary-tree/description 480.https://www…

RemoveDuplicatesFromSortedArrayI: 问题描述:给定一个有序数组,去掉其中重复的元素.并返回新数组的长度.不能使用新的空间. [1,1,2,3] -> [1,2,3] 3 算法思路:用一个数字记录新数组的最后一个元素的位置 public class RemoveDuplicatesFromSortedArray { public int removeDuplicates(int[] nums) { if(nums.length == 0) { return 0; }…

/* * @lc app=leetcode.cn id=26 lang=c * * [26] 删除排序数组中的重复项 * * https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array/description/ * * algorithms * Easy (42.77%) * Total Accepted: 89.1K * Total Submissions: 208.1K * Testcase Example: '[…

1. 原题链接 https://leetcode.com/problems/remove-duplicates-from-sorted-array/description/ 2. 题目要求 给定一个已经排序的整数数组nums[ ],返回除去重复元素后的数组长度 注意:不能重新创建一个数组,空间复杂度为O(1) 3. 解题思路 使用指针j来遍历数组,i用来计数.初始时,i指向nums[0],j指向nums[1]. 当nums[i] != nums[j]时,i++,且nums[i]=nums[j],…

任务二:删除排序数组中的重复项 原文链接:https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array/ 最开始的解决思路: 遍历2次数组: 第一次,遍历整个数组,放置标记index1 第二次,从每一个第一次遍历的元素index1后面开始遍历index2. 然后,一个一个比较,重复的元素,删除后来再次重复的元素index2. for index1 in range(0, len( num_list) - 1): for…