就是借用字典的key-value进行的匹配,算法复杂度仅为O(n) 然后还看到了这个 1classSolution(object):2deftwoSum(self, nums, target):34forind, numinenumerate(nums):5iftarget-numinnumsandnums.index(target-num) !=ind:6return[ind, nums.index(target-num)]7return-1 直接利用了列表的索引。 运行速度...
这是Problem 1的反问题,但这一个题难度要稍微大一些,我们考虑一下怎么做。 这里我们也是考虑使用两个栈来求解。区分于队列,栈是先进后出的,因此每做一次栈的操作都会改变元素的顺序。那么假如说我们有两个栈stack1, stack2。那么一开始我们输入数据的时候,可以把它放到stack1里面,只要我们需要输出了,我们就看看s...
As can be seen from the above process, the problem of brute force cracking is that the time complexity ofis too high, and the reason is high becausehas some skipped processesin the traversal process. In order to facilitate understanding, we bring in a problem Example 1 in the case oftarget...
classSolution{ public: vector<int>twoSum(vector<int>&nums,inttarget) { } }; 已存储 行1,列 1 运行和提交代码需要登录 Case 1Case 2Case 3 nums = [2,7,11,15] target = 9 9 1 2 3 4 5 6 › [2,7,11,15] 9 [3,2,4] ...
首先，将1~n的数字初始化到hashSet里，然后判断原数组nums的元素如果在hashSset里面，则移除，最后剩下...
本文主要介绍作者对最长回文子串问题( Longest Palindromic Substring)的动态规划解法(DP solution),作者leetcode上的提交结果为accept。作者之前也采用了DP算法解决这个问题,但多次都不能accept,提交结果为TLE(time limit exceeded)。作者经过多次修改才成功,所以希望把自己的成功代码贡献出来。作者的失败经验与反思见后。
2 3 4 5 6 classSolution{ public: intmaxArea(vector<int>&height) { } }; 已存储 行1,列 1 运行和提交代码需要登录 Case 1Case 2 height = [1,8,6,2,5,4,8,3,7] 9 1 2 › [1,8,6,2,5,4,8,3,7] [1,1] Source
solution.h the skyline problem Sep 8, 2015 Repository files navigation README Binary Tree Postorder Traversal (145) 要求不用递归实现后序遍历 后序是left-right-root,那么首先用修正的前序root-right-left,然后reverse一下,变成left-right-root就行了,代码如下:Factorial...
class Solution { public: int sumDistance(vector<int>& nums, string s, int d) { vect...
// Solution 2: This is a clever solution. (I'm not saying it's a short solution.) Took me more than an hour to figure out. First of all, I believed there's a better solution to this problem than an O(N ^ 2) brute-force one. Because it's about "order", and you know there...