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leetcode 238. Product of Array Except Self (乱搞)

·411 字·1 分钟
ACM Leetcode

Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

Follow up: Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)

先来个O(n)空间的无脑解法。。。一个前缀积一个后缀积就好了。。。

 1class Solution {
 2public:
 3    vector<int> productExceptSelf(vector<int>& nums) {
 4        vector<int>res;
 5        vector<int>pre,suf;
 6        int siz = nums.size();
 7        if (siz==0) return res;
 8        for ( int i = 0 ; i < siz ; i++)
 9        {
10            int tmp;
11            if (i==0) tmp = nums[0];
12            else tmp = pre[i-1]*nums[i];
13            pre.push_back(tmp);
14        }
15        int cnt = 0 ;
16        for ( int i = siz -1 ; i>= 0 ; i--)
17        {
18            int tmp;
19            if (i==siz-1) tmp = nums[siz-1];
20            else tmp = suf[cnt++]*nums[i];
21            suf.push_back(tmp);
22        }
23      //  for ( int i = 0 ; i < siz ; i++) printf("%d ",pre[i]); puts("");
24    //    for ( int i = 0 ; i < siz ; i++) printf("%d ",suf[i]);puts("");
25        reverse(suf.begin(),suf.end());
26        for ( int i = 0 ; i < siz;  i++)
27        {
28            int l = i-1;
29            int r = i+1;
30            int x,y;
31            if (l<0) x = 1;
32            else x = pre[l];
33            if (r>=siz) y = 1;
34            else y = suf[r];
35      //      cout<<"x:"<<x<<" y:"<<y<<endl;
36            res.push_back(x*y);
37        }
38        return res;
39
40    }
41};

常数空间的做法。。。想了半天没有想法。。。

看了solution…发现就是借用res数组搞事情。。。。每个位置的答案扫完两遍得到。。。

感觉。。。非常的。。。无聊啊。。。。真心没意思吧，还以为是什么巧妙的做法呢。。。

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leetcode 79. Word Search (dfs)

·374 字·1 分钟
ACM Leetcode DFS
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. The same letter cell may not be used more than once. 思路：dfs即可。记得要回溯一下… 代码实现 1/* *********************************************** 2Author :111qqz 3Created Time :2017年04月07日 星期五 14时32分54秒 4File Name :79.cpp 5************************************************ */ 6class Solution { 7 8public: 9 10 int n,m; 11 const int dx4[4]={1,-1,0,0}; 12 const int dy4[4]={0,0,-1,1}; 13 bool vis[1005][1005]; 14 int len; 15 16 bool yes( int x,int y) 17 { 18 if (x>=0&&x<=n-1&&y>=0&&y<=m-1) return true; 19 return false; 20 } 21 22 bool dfs( int x,int y,int cur,vector<vector<char> > & maze,string & st) 23 { 24// printf("x:%d y:%d cur : %d\n",x,y,cur); 25 if (cur>=len) return true; 26 for ( int i = 0 ; i < 4 ; i++) 27 { 28 int nx = x + dx4[i]; 29 int ny = y + dy4[i]; 30 if (!yes(nx,ny)) continue; 31 if (vis[nx][ny]) continue; 32 if (maze[nx][ny]!=st[cur]) continue; 33 vis[nx][ny] = true; 34 bool res = dfs(nx,ny,cur+1,maze,st); 35 if (res) return true; 36 vis[nx][ny] = false ;//好像还要回溯一下啊？ 37 } 38 return false; 39 } 40 41 bool exist(vector<vector<char>>& board, string word) { 42 n = board.size(); 43 if (n==0) return false; 44 m = board[0].size(); 45 if (m==0) return false; 46 len = word.length(); 47 cout<<"len:"<<len<<endl; 48 char beg = word[0]; 49 for ( int i = 0 ; i < n ; i++) 50 for ( int j = 0 ; j < m ; j++) 51 if (board[i][j]==beg) 52 { 53 memset(vis,false,sizeof(vis)); 54 //起点忘记打标记了。。。智力-2.。。 55 vis[i][j] = true; 56 bool ok = dfs(i,j,1,board,word); 57 cout<<"ok:"<<ok<<endl; 58 if (ok) return true; 59 } 60 return false; 61 62 63 } 64 65};

leetcode 80 Remove Duplicates from Sorted Array II （有序数组去除重复元素）

·324 字·1 分钟
ACM Leetcode
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leetcode 81. Search in Rotated Sorted Array II (有重复元素的旋转数组找给定值)

·333 字·1 分钟
ACM 二分 Leetcode
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). Write a function to determine if a given target is in the array. The array may contain duplicates. 好像阿里一面的时候问过。。。 思路：肯定是二分。。。不过由于有重复元素。。。所以很恶心。。。

leetcode 289. Game of Life (模拟)

·1065 字·3 分钟
ACM Leetcode
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leetcode 90. Subsets II (枚举子集)

·266 字·1 分钟
ACM 枚举子集 Leetcode
Given a collection of integers that might contain duplicates, nums, return all possible subsets. Note: The solution set must not contain duplicate subsets. For example, If nums = [1,2,2], a solution is: 1[ 2 [2], 3 [1], 4 [1,2,2], 5 [2,2], 6 [1,2], 7 [] 8] 思路： 复习（？）一下 枚举子集的三种写法 （还有种更飘逸的…先不写了orz 这道题我用位向量法A的。。