C++实现LeetCode(80.有序数组中去除重复项之二)

[LeetCode] 80. Remove Duplicates from Sorted Array II 有序数组中去除重复项之二

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,1,2,2,3],

Your function should return length =

5

, with the first five elements of

nums

being

1, 1, 2, 2

and 3 respectively.

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,1,2,3,3],

Your function should return length =

7

, with the first seven elements of

nums

being modified to 

0

, 0, 1, 1, 2, 3 and 3 respectively.

It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}

这道题是之前那道 Remove Duplicates from Sorted Array 的拓展,这里允许最多重复的次数是两次,那么可以用一个变量 cnt 来记录还允许有几次重复,cnt 初始化为1,如果出现过一次重复,则 cnt 递减1,那么下次再出现重复,快指针直接前进一步,如果这时候不是重复的,则 cnt 恢复1,由于整个数组是有序的,所以一旦出现不重复的数,则一定比这个数大,此数之后不会再有重复项。理清了上面的思路,则代码很好写了:

解法一:

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int pre = 0, cur = 1, cnt = 1, n = nums.size();
        while (cur < n) {
            if (nums[pre] == nums[cur] && cnt == 0) ++cur;
            else {
                if (nums[pre] == nums[cur]) --cnt;
                else cnt = 1;
                nums[++pre] = nums[cur++];
            }
        }
        return nums.empty() ? 0 : pre + 1;
    }
};

这里其实也可以用类似于 Remove Duplicates from Sorted Array 中的解法三的模版,由于这里最多允许两次重复,那么当前的数字 num 只要跟上上个覆盖位置的数字 nusm[i-2] 比较,若 num 较大,则绝不会出现第三个重复数字(前提是数组是有序的),这样的话根本不需要管 nums[i-1] 是否重复,只要将重复个数控制在2个以内就可以了,参见代码如下:

解法二:

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int i = 0;
        for (int num : nums) {
            if (i < 2 || num > nums[i - 2]) {
                nums[i++] = num;
            }
        }
        return i;
    }
};

到此这篇关于C++实现LeetCode(80.有序数组中去除重复项之二)的文章就介绍到这了,更多相关C++实现有序数组中去除重复项之二内容请搜索我们以前的文章或继续浏览下面的相关文章希望大家以后多多支持我们!

(0)

相关推荐

  • C++实现LeetCode(82.移除有序链表中的重复项之二)

    [LeetCode] 82. Remove Duplicates from Sorted List II 移除有序链表中的重复项之二 Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Example 1: Input: 1->2->3->3->4->4->5 Outp

  • C++实现LeetCode(75.颜色排序)

    [LeetCode] 75. Sort Colors 颜色排序 Given an array with n objects colored red, white or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white and blue. Here, we will use the integers 0, 1, and 2

  • C++实现LeetCode(76.最小窗口子串)

    [LeetCode] 76. Minimum Window Substring 最小窗口子串 Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). Example: Input: S = "ADOBECODEBANC", T = "ABC" Output: "BA

  • C++实现LeetCode(74.搜索一个二维矩阵)

    [LeetCode] 74. Search a 2D Matrix 搜索一个二维矩阵 Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted from left to right. The first integer of each row is great

  • C++实现LeetCode(73.矩阵赋零)

    [LeetCode] 73.Set Matrix Zeroes 矩阵赋零 Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place. click to show follow up. Follow up: Did you use extra space? A straight forward solution using O(mn) space is probably

  • C++实现LeetCode(81.在旋转有序数组中搜索之二)

    [LeetCode] 81. Search in Rotated Sorted Array II 在旋转有序数组中搜索之二 Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]). You are given a target value to search.

  • C++实现LeetCode(72.编辑距离)

    [LeetCode] 72. Edit Distance 编辑距离 Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2. You have the following 3 operations permitted on a word: Insert a character Delete a character Replace a char

  • C++实现LeetCode(79.词语搜索)

    [LeetCode] 79. Word Search 词语搜索 Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. Th

  • C++实现LeetCode(80.有序数组中去除重复项之二)

    [LeetCode] 80. Remove Duplicates from Sorted Array II 有序数组中去除重复项之二 Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length. Do not allocate extra space for another array, you mus

  • C++实现LeetCode(26.有序数组中去除重复项)

    [LeetCode] 26. Remove Duplicates from Sorted Array 有序数组中去除重复项 Given a sorted array nums, 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

  • java算法入门之有效的括号删除有序数组中的重复项实现strStr

    目录 1.LeetCode 20.有效的括号 题目 小编菜解 思路及算法 大神解法 2.LeetCode 26.删除有序数组中的重复项 题目 小编菜解初版 小编菜解改进版 思路及算法 大神解法 3.LeetCode 28.实现strStr 题目 小编菜解 大神解法 也许,我们永远都不会知道自己能走到何方,遇见何人,最后会变成什么样的人,但一定要记住,能让自己登高的,永远不是别人的肩膀,而是挑灯夜战的自己,人生的道路刚刚启程,当你累了倦了也不要迷茫,回头看一看,你早已不再是那个年少轻狂的少年. 1

  • C语言 详解如何删除有序数组中的重复项

    目录 删除有序数组中的重复项Ⅰ a.思路 b.图解 c.代码 d.思考 删除有序数组中的重复项Ⅱ a.思路 b.图解 c.代码 d.思考 删除有序数组中的重复项Ⅰ a.思路 定义变量 int dest=0,cur=1,nums[cur]与nums[dest]逐一比较. nums[cur]!=nums[dest],将nums[cur]放入dest下一个位置,更新dest. nums[cur]!=nums[dest],cur移动. cur==numsSize,结束.返回dest+1. b.图解 c.

  • JavaScript删除有序数组中的重复项

    目录 如果有一个有序数组 nums ,要求原地 删除重复出现的元素,使每个元素 只出现一次 ,返回删除后数组的新长度. 不要使用额外的数组空间,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. 说明: 为什么返回数值是整数,但输出的答案是数组呢? 注意:输入数组是以「引用」方式传递的,这意味着在函数里修改输入数组对于调用者是可见的. 你可以想象内部操作如下: // nums 是以"引用"方式传递的.也就是说,不对实参做任何拷贝 int len = removeDu

  • js数组中去除重复值的几种方法

    在日常开发中,我们可能会遇到将一个数组中里面的重复值去除,那么,我就将我自己所学习到的几种方法分享出来 去除数组重复值方法: 1,利用indexOf()方法去除 思路:创建一个新数组,然后循环要去重的数组,然后用新数组去找要去重数组的值,如果找不到则使用.push添加到新数组,最后把新数组返回回去就行了 看不懂没关系,上代码就比较容易懂了 function fun(arr){ let newsArr = []; for (let i = 0; i < arr.length; i++) { if(

  • C++实现LeetCode(83.移除有序链表中的重复项)

    [LeetCode] 83. Remove Duplicates from Sorted List 移除有序链表中的重复项 Given a sorted linked list, delete all duplicates such that each element appear only once. Example 1: Input: 1->1->2 Output: 1->2 Example 2: Input: 1->1->2->3->3 Output: 1-

  • JS实现合并两个数组并去除重复项只留一个的方法

    本文实例讲述了JS实现合并两个数组并去除重复项只留一个的方法.分享给大家供大家参考,具体如下: //It's merge arr1 and arr2 , delete the same element only leave one //It's only apdapter array. If object, no. //The sequence of the two array is not required. mergeArray:function (arr1, arr2){ for (var

  • PHP二维数组实现去除重复项的方法【保留各个键值】

    本文实例讲述了PHP二维数组实现去除重复项的方法.分享给大家供大家参考,具体如下: 对于如下二维数组,要求对其进行去重: $arr = array( '0'=>array( 'name'=>'james', 'age'=>30, ), '1'=>array( 'name'=>'susu', 'age'=>26, ), '2'=>array( 'name'=>'james', 'age'=>30, ), 'new'=>array( 'name'=&

随机推荐