Leetcode

本来不知道写什么所以不打算写了，不过后来觉得可以把今年做的一些重大的决定写出来，把当时的分析和想法记录下来。这样若干年后再回看，就能找到：是哪些明智或愚蠢的决定对人生产生了巨大的影响。

请实现最近最少使用缓存(Least Recently Used (LRU) cache)类，需要支持 get, set，操作。 get 操作，给出 key，获取到相应的 value (value 为非负数)，如果不存在返回-1, 如果存在此 key 算作被访问过。 set 操作，设置 key，如果 key 存在则覆盖之前的 value (此时相当于访问过一次)。 如果 key 不存在，需要进行插入操作，如果此时已经 key 的数量已经到达 capacity, 这样需要淘汰掉最近最少使用(也就是上次被使用的时间距离现在最久的)的那 一项。

A peak element is an element that is greater than its neighbors. Given an input array where num[i] ≠ num[i+1], find a peak element and return its index. The array may contain multiple peaks, in that case return the index to any one of the peaks is fine. You may imagine that num[-1] = num[n] = -∞. For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.

Find the contiguous subarray within an array (containing at least one number) which has the largest product. For example, given the array [2,3,-2,4], the contiguous subarray [2,3] has the largest product = 6. 思路：由于有正，有负，还有0.。。所以比最大子串之和要复杂一些。。。 dp[i].max表示到当前位置的最大乘积。

Given a sorted integer array without duplicates, return the summary of its ranges. For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"]. 题意：把连续的数连续表示 思路：模拟。注意有负数，注意有-2147483648这种数据。 本来还想着，可能是leetcode加数据的审核机制太松，导致被人加了奇怪的数据。。。

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn’t one, return 0 instead. For example, given the array [2,3,1,2,4,3] and s = 7, the subarray [4,3] has the minimal length under the problem constraint 思路：尺取即可。。好久没写，竟然调了半天。。。 代码实现 1/* *********************************************** 2Author :111qqz 3Created Time :2017年04月13日 星期四 20时48分00秒 4File Name :209.cpp 5************************************************ */ 6class Solution { 7 8public: 9 10 int ruler(vector<int>nums,int tar,int n) 11 { 12 int head = 0; 13 int tail = 0; 14 int sum = 0 ; 15 int res = 0x3f3f3f3f; 16 while (tail<n&&head<=tail) 17 { 18 sum = sum + nums[tail]; 19 if (sum>=tar) 20 { 21 res = min(res,tail-head+1); 22 while (sum>=tar&&head<tail) 23 { 24 sum-=nums[head]; 25 head++; 26 } 27 if (sum>=tar) 28 { 29 res = min(res,tail-head+1); 30 } 31 else 32 { 33 head--; 34 sum+=nums[head]; 35 res = min(res,tail-head+1); 36 } 37 } 38 39 40 41 tail++; 42 } 43 return res==0x3f3f3f3f?0:res; 44 } 45 46 47 int minSubArrayLen(int s, vector<int>& nums) { 48 int n = nums.size(); 49 int res = ruler(nums,s,n); 50 return res; 51 52 53 } 54 55};

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space. 题意：给你n个数，要求找出出现此处大于n/3的。。。 思路：之前做过一个找出n个数出现此处大于n/2的题目，思想是“非吾族类，其心必异”。。

Given an array with n objects colored red, white or blue, sort them 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 to represent the color red, white, and blue respectively. 题意：一个数组，由0,1,2组成，现在要求升序排列 思路：无脑做法就是计数排序，扫两遍，时间复杂度O(n)，空间复杂度O(1)

Given n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water. Note: You may not slant the container and n is at least 2. 题意：n条竖直的线段 (i,0)->(i,a[i])，从中选2条，和x轴共同组成一个开口的容器，问容器的最大面积。

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution. 思路： 排序，然后two pointer，复杂度 O(n^2) 1/* *********************************************** 2Author :111qqz 3Created Time :2017年04月13日 星期四 16时24分28秒 4File Name :16.cpp 5************************************************ */ 6class Solution { 7 8public: 9 10 int n; 11 int threeSumClosest(vector<int>& nums, int target) { 12 n = nums.size(); 13 sort(nums.begin(),nums.end()); 14 int mn = 0x3f3f3f3f; 15 int x,y,z; 16 for ( int i = 0 ; i <=n-3 ; i++) 17 { 18 int head = i+1; 19 int tail = n-1; 20 int tar = target - nums[i]; 21 while (head<tail) 22 { 23 int cur = target-nums[i]-nums[head]-nums[tail]; 24 if (abs(cur)<mn) 25 { 26 mn = abs(cur); 27 x = nums[i]; 28 y = nums[head]; 29 z = nums[tail]; 30 } 31 if (nums[head]+nums[tail]==tar) return target; 32 if (nums[head]+nums[tail]<tar) head++; 33 else tail--; 34 } 35 } 36 return x + y + z; 37 38 39 40 41 } 42 43};

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. Note: The solution set must not contain duplicate quadruplets. 思路： O(n^2)枚举两个元素，变成2-sum问题，总体复杂度O(n^3)

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero. Note: The solution set must not contain duplicate triplets. 思路：排序O(nlgn)，然后枚举一个元素O(n)，对于每个元素，在剩下的区间中 two pointer O(n)

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers. 题意：1..9个数，从中选择k个，和为n，要求输出所有满足题意的集合。 思路：枚举子集，根据sum和集合元素个数剪枝即可。

Given two integers n and k, return all possible combinations of k numbers out of 1 … n. 思路：就是枚举子集，根据集合的大小剪枝。。。最后只要集合大小为k的集合 1/* *********************************************** 2Author :111qqz 3Created Time :2017年04月13日 星期四 15时25分37秒 4File Name :77.cpp 5************************************************ */ 6class Solution { 7public: 8 set<vector<int> >se; 9 int B[1005]; 10 vector<vector<int> >res; 11 void get_subset(int n,int *B,int cur,int cnt,int k) 12 { 13 if (cur==n) 14 { 15 vector<int>tmp; 16 for ( int i = 0 ; i < n ; i++) 17 if (B[i]) tmp.push_back(i+1); 18 if (tmp.size()==k) 19 se.insert(tmp); 20 return ; 21 } 22 if (cnt<k) 23 { 24 B[cur] = 1; 25 get_subset(n,B,cur+1,cnt+1,k); 26 } 27 B[cur] = 0 ; 28 get_subset(n,B,cur+1,cnt,k); 29 } 30 vector<vector<int>> combine(int n, int k) { 31 get_subset(n,B,0,0,k); 32 for (auto &it:se) 33 { 34 res.push_back(it); 35 } 36 return res; 37 } 38};

The set [1,2,3,…,_n_] contains a total of n! unique permutations. By listing and labeling all of the permutations in order, We get the following sequence (ie, for n = 3): 1. `"123"` 2. `"132"` 3. `"213"` 4. `"231"` 5. `"312"` 6. `"321"` Given n and k, return the _k_th permutation sequence. Note: Given n will be between 1 and 9 inclusive. 思路：还是根据leetcode 31 解题报告 中的算法搞一下就好了。。

Given a collection of numbers that might contain duplicates, return all possible unique permutations.__ 思路：和leet code 46 类似，最后用set去个重即可。。 代码实现 1/* *********************************************** 2Author :111qqz 3Created Time :2017年04月13日 星期四 15时00分48秒 4File Name :47.cpp 5************************************************ */ 6class Solution { 7 8public: 9 void solve( vector<int>&nums) 10 { 11 int n = nums.size(); 12 if (n==0) return; 13 int k = -1; 14 for ( int i = n-2 ; i >= 0 ; i--) 15 { 16 if (nums[i]<nums[i+1]) 17 { 18 k = i; 19 break; 20 } 21 } 22 if (k==-1) 23 { 24 reverse(nums.begin(),nums.end()); 25 return; 26 } 27 int l = -1; 28 for ( int i = n-1 ; i >k ; i--) 29 { 30 if (nums[k]<nums[i]) 31 { 32 l = i; 33 break; 34 } 35 } 36 swap(nums[l],nums[k]); 37 reverse(nums.begin()+k+1,nums.end()); 38 } 39 40 void pr (vector<int> &nums) 41 { 42 int siz = nums.size(); 43 for ( int i = 0 ; i < siz; i++) 44 printf("%d%c",nums[i],i==siz-1?'\n':' '); 45 } 46 vector<vector<int>> permuteUnique(vector<int>& nums) { 47 set<vector<int> >se; 48 vector<vector<int> >res; 49 int n = nums.size(); 50 int total = 1 ; 51 for ( int i = 2 ; i <= n ; i++) total*=i; 52 53 for ( int i = 1 ; i <= total ; i++) 54 { 55 se.insert(nums); 56// pr(nums); 57 solve(nums); 58 } 59 for ( auto &it :se) 60 { 61 res.push_back(it); 62 } 63 return res; 64 } 65 66};

Given a collection of distinct numbers, return all possible permutations. 思路：调用n-1次 leetcode 31 解题报告 中提到的算法即可。。。 代码实现 1/* *********************************************** 2Author :111qqz 3Created Time :2017年04月13日 星期四 14时49分34秒 4File Name :46.cpp 5 ************************************************ */ 6class Solution { 7 8 public: 9 10 void solve(vector<int> &nums) 11 { 12 int n = nums.size(); 13 if (n==0) return; 14 int k = -1; 15 for ( int i = n-2 ; i >= 0 ; i--) 16 { 17 if (nums[i]<nums[i+1]) 18 { 19 k = i ; 20 break; 21 } 22 } 23 if (k==-1) 24 { 25 reverse(nums.begin(),nums.end()); 26 return; 27 } 28 int l = -1; 29 for ( int i = n-1 ; i > k ; i-- ) 30 { 31 if (nums[k]<nums[i]) 32 { 33 l = i ; 34 break; 35 } 36 } 37 swap(nums[l],nums[k]); 38 reverse(nums.begin()+k+1,nums.end()); 39 } 40 vector<vector<int>> permute(vector<int>& nums) { 41 vector<vector<int> >res; 42 int n = nums.size(); 43 int total = 1; 44 for ( int i = 2 ; i <= n ; i++) total*=i; 45 for ( int i = 1 ; i <= total; i++) 46 { 47 res.push_back(nums); 48 solve(nums); 49 } 50 51 return res; 52 } 53 54};

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). The replacement must be in-place, do not allocate extra memory. Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column. 1,2,3 → 1,3,2 3,2,1 → 1,2,3 1,1,5 → 1,5,1 思路： 参考了wiki_Permutation

Suppose an array sorted in ascending order 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, otherwise return -1. You may assume no duplicate exists in the array. 思路：找规律。。。二分。。。 0 1 2 3 4 5 6 1 2 3 4 5 6 0 2 3 4 5 6 0 1 3 4 5 6 0 1 2 4 5 6 0 1 2 3 5 6 0 1 2 3 4 6 0 1 2 3 4 5 观察发现。。。a[mid]<a[r]的时候，后半段有序；

Given an array of integers sorted in ascending order, find the starting and ending position of a given target value. Your algorithm’s runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1]. For example, Given [5, 7, 7, 8, 8, 10] and target value 8, return [3, 4]. 思路：二分。。。 我好像根本不会二分啊？？？ 二分查找