Leetcode(4)-Two Sum

01.Two Sum

//LeetCode, Two Sum hash存储每个数字的下标class Solution {public:vector<int> twoSum(vector<int> &nums, int target) {    unordered_map<int, int> mapping;    vector<int> result;    for (int i = 0; i < nums.size(); i++) {        mapping[nums[i]] = i;    }    for (int i = 0; i < nums.size(); i++) {        int gap = target - nums[i];        if (mapping.find(gap) != mapping.end() && mapping[gap] > i) {            result.push_back(i + 1);            result.push_back(mapping[gap] + 1);            break;        }    }    return result;    }};

15 3sum
先升序排序，然后用第一重for循环确定第一个数字。
然后在第二重循环里，第二、第三个数字分别从两端往中间扫。
如果三个数的sum等于0，得到一组解。
如果三个数的sum小于0，说明需要增大，所以第二个数往右移。
如果三个数的sum大于0，说明需要减小，所以第三个数往左移。
注意要跳过相同元素，否则也会超时；

class Solution {public:    vector<vector<int>> threeSum(vector<int>& nums) {        vector<vector<int> > ret;        int n = nums.size();        sort(nums.begin(), nums.end());        int sum = 0;        int i,j,k;        for(i=0; i<n; i++){            //skip same i            while(i > 0 && i < n && nums[i] == nums[i-1])                i ++;            j=i+1;k=n-1;            while(j<k){                sum = nums[i]+nums[j]+nums[k];                if(sum==0){                    vector<int> cur(3);                    cur[0] = nums[i];                    cur[1] = nums[j];                    cur[2] = nums[k];                    ret.push_back(cur);                    j++; k--;                    //skip same j                    while(j < k && nums[j] == nums[j-1])                        j ++;                    //skip same k                    while(k > j && nums[k] == nums[k+1])                        k --;                }                else if(sum>0){                    //sum大于0，说明需要减小，所以第三个数往左移                    k--;                    //skip same k                    while(k > j && nums[k] == nums[k+1])                        k--;                }                else{                    //sum小于0，说明需要增大，所以第二个数往右移。                    j++;                    //skip same j                    while(j < k && nums[j] == nums[j-1])                        j++;                }            }        }        return ret;    }};

18 4sum
与3sum思路一致，也是扫描，此处使用三重循环；
注意去重；
[-3,-2,-1,0,0,1,2,3]
0
Output:
[[-3,-2,2,3],[-3,-1,1,3],[-3,0,0,3],[-3,0,1,2],[-3,0,1,2],[-2,-1,0,3],[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
Expected:
[[-3,-2,2,3],[-3,-1,1,3],[-3,0,0,3],[-3,0,1,2],[-2,-1,0,3],[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]

class Solution {public:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int> > ret;        //去重！！！        set<vector<int>> res;        int n = nums.size();        if(n<4) return ret;        sort(nums.begin(), nums.end());        int sum = 0;        int i,j,k,h;        for(i=0; i<n-2; i++){            //skip same i            while(i > 0 && i < n && nums[i] == nums[i-1])                i ++;            for(j=i+1; j<n-1; j++){                k=j+1;h=n-1;                while(k<h){                    sum = nums[i]+nums[j]+nums[k]+nums[h];                    if(sum==target){                        vector<int> cur(4);                        cur[0] = nums[i];                        cur[1] = nums[j];                        cur[2] = nums[k];                        cur[3] = nums[h];                        //集合是insert不是push_back了;                        res.insert(cur);                        k++; h--;                        while(k < h && nums[k] == nums[k-1])                            k++;                        while(h > k && nums[h] == nums[h+1])                            h--;                    }                    else if(sum>target){                        h--;                        while(h > k && nums[h] == nums[h+1])                            h--;                    }                    else{                        k++;                        while(k < h && nums[k] == nums[k-1])                            k++;                    }                }            }        }        //格式化        set<vector<int>>::iterator it = res.begin();            for(; it != res.end(); it++)                ret.push_back(*it);           return ret;    }};

79、word search
先要找到’首字母’,之后按走迷宫的方式探索以达到正确目标；..递归过程要改改、、、

class Solution {public:    //获取邻居位置    vector<int> istheneighbor(vector<vector<char>>& board, vector<int> a, char b){        int x = a[0];        int y = a[1];        vector<int> t;        if((x+1<board.size()) && (board[x+1][y] == b))        {             t.push_back(x+1);             t.push_back(y);        }        if((y+1<board[0].size()) && (board[x][y+1] == b))        {             t.push_back(x);             t.push_back(y+1);        }        if((x-1>0) && (board[x-1][y] == b))        {             t.push_back(x-1);             t.push_back(y);        }        if((y-1>0) && (board[x][y-1] == b))        {             t.push_back(x);             t.push_back(y-1);        }        return t;    }    //获取首字母所在所有位置的列表    vector<vector<int>> getalp(vector<vector<char>>& board, char a){        vector<vector<int>>path;        vector<int> t;        for(int i=0; i< board.size(); i++)        {            for(int j=0; j < board[0].size(); j++)            {                if(board[i][j] == a)                {                    t.push_back(i);                    t.push_back(j);                    path.push_back(t);                }            }        }        return path;    }    //递归    bool dp(vector<vector<char>>& board,vector<int> t,string word,int num){        if(num == word.size()-1) return true;        num++;        vector<int> x = istheneighbor(board,t,num);        if(x.size()==2)             dp(board,x,word,num);        else             return false;    }    bool exist(vector<vector<char>>& board, string word) {        vector<vector<int>> alp = getalp(board, word[0]);        for(int i=0;i<alp.size();i++)        {            bool f = dp(board,alp[i],word,0);            if(f == true)                return true;        }        return false;    }};