lessfish/leetcode

【NO LONGER UPDATE】:pencil2: LeetCode solutions with JavaScript

leetcode's Issues

nice to meet U--just want to say hello

hi this is my first time to write an issue.

i am from cnblogs 小刺猬001.

that‘s all. thanks for your reading~

wish you have a good mood today.

同行

你好，我之前找工作也是用的js刷的题，大概刷了240多题， 如果有需要我也可以submit我的一些code

很好的repo

对了，我最近也在静心研读underscore~~有机会可以交流交流
我的email:[email protected]

thx

test issues

test

3Sum.js

The javascript version is exceeding time limit for longer list.
Perhaps you may refer to following java code to develop a javascript version.

    public List<List<Integer>> threeSum(int[] nums) {
        
        List<Integer> list;
        List<List<Integer>> retList = new  LinkedList<>();
        Arrays.sort(nums);
        for( int i = 0; i < nums.length && nums[i] <= 0; i++){
            if( i > 0 && nums[i] == nums[i-1])
                continue;
            int sum = (-1)*nums[i], front = i + 1, behin = nums.length - 1;
            while( front < behin )
                if( nums[front] + nums[behin] > sum) 
                    behin--;
                else if( nums[front] + nums[behin] < sum) 
                    front++;
                else{
                    list = new LinkedList<Integer>();
                    list.add(nums[i]);
                    list.add(nums[front]);
                    list.add(nums[behin]);
                    retList.add(list);
                    while( front<behin && nums[front]==nums[front+1]) front++;
                    while( front<behin && nums[behin]==nums[behin-1]) behin--;
                    front++;
                    behin--;
                }
        }
        return retList;
    }

Valid Parenthesis

The golang solution using byte to validate the parenthesis hits 0ms compared to ~108ms in javascript.
Maybe we shall try to find a better solution.

func isValid(s string) bool {
	if len(s)%2 != 0 {
		return false
	}
	left := map[byte]struct{}{
		'(': struct{}{},
		'[': struct{}{},
		'{': struct{}{},
	}
	right := map[byte]struct{}{
		')': struct{}{},
		']': struct{}{},
		'}': struct{}{},
	}
	bytes := []byte(s)
	stack := []byte{}
	for i := range bytes {
		if _, ok := left[bytes[i]]; ok {
			stack = append(stack, bytes[i])
		} else if _, ok = right[bytes[i]]; ok {
			if len(stack) == 0 {
				return false
			}
			switch stack[len(stack)-1] {
			case '(':
				if bytes[i] != ')' {
					return false
				}
			case '[':
				if bytes[i] != ']' {
					return false
				}
			case '{':
				if bytes[i] != '}' {
					return false
				}
			}
			stack = stack[:len(stack)-1]
		} else {
			return false
		}
	}
	if len(stack) == 0 {
		return true
	}
	return false
}

关于readme.md的 update

你好，从博客园一路逛过来看的。感觉 leetcode 上能拿 js 刷了这么多题的人真不多见😂，我之前也是用 JS， 后来还是随大流用了 Java

主要想问下你 readme.md 是写了什么脚本更新吗？方便就此交流一下吗？
谢谢

试了下你的 generate.js 爬出貌似失效了

$('#problemList tbody tr') 是空

去网站上搜 problemList 也没有。

是leetcode更新了页面dom，还是我打开方式不对 :-(

The solution of "Reverse Words in a String" is not correct

I think you misunderstand the leetcode description about this problem
That's my test case about this problem, maybe will help you to check your solution
TestCase

Add Two Numbers这个算法有什么用啊？

看样子就是把链表的每一个数字加起来，如果超过10就取余数，进的一个1就参与到后面2个数的相加里去，最后得出一个新的链表，这样做的意义是什么？哪些实际问题是需要这种算法的？

Find peak element problem can be solved by Binary Search Algorithm

Currently, it is solved in sequential approach which adds time complexity of O(n) but it can be solved using binary search method which eventually results in better time complexity of O(log n).

using constant space complexity.

Sort List 这题忽略了 using constant space complexity.

链表转数组虽然ac很快，但链表本身失去了意义。

Interested in contributing

@hanzichi 哈喽，我最近自己也在JS写LeetCode，想要contribute，不知道有没有什么要求呢？