Leetcode: Array

1. Longest-consecutive-sequence

Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

For example,
GIVEN[100, 4, 200, 1, 3, 2],
The longest consecutive elements sequence is[1, 2, 3, 4]. Return its length:4.

Your algorithm should run in O (n) complexity.

Given an array of shapes, the longest sequential sequence is obtained. For example, the array [100,4,200,1,3,2], the longest continuous sequence length is [1,2,3,4], the length is 4. Requires a time complexity of O (n).

Sort words to at least O (nlgn) complexity. O (n) complexity, currently only found using hash to solve the scheme, add, remove, contains and other methods of the complexity is O (1), so two times the operation of the traversal is O (n).

 Public Static intLongestconsecutive (int[] num) {    //if array is empty, return 0    if(Num.length = = 0) {        return0; } Set<Integer> set =NewHashset<integer>(); intmax = 1;  for(inte:num) Set.add (e);  for(inte:num) {        intleft = E-1; intright = e + 1; intCount = 1;  while(Set.contains (left)) {count++;            Set.remove (left); Left--; }          while(Set.contains (right)) {count++;            Set.remove (right); Right++; } Max=Math.max (count, Max); }     returnMax;}

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2. Surrounded-regions

Given a 2D board containing ' x ' and ' O ', capture all regions surrounded by ' x '.

A region was captured by flipping all ' O ' s into the ' X ' s in this surrounded region.

For example,

x x x xx o o xx x o xx o x x

After running your function, the board should is:

x x x xx x x xx x x xx O x x

A typical BFS problem. Traverse each character, if "O", start the BFS traversal from the current character, if the surrounding is also "O" to join the current traversal of the queue, know to traverse all adjacent "O", at the same time, to determine whether each o is surrounded, only by an O is not surrounded, The currently traversed set of O is not surrounded, because these o are all connected.

Of course, this problem can also be used with DFS, only the test data is too large, submitted StackOverflow.

1) BFS Breadth First Search

 Public classSolution {//Use a queue to do BFS    Privatequeue<integer> queue =NewLinkedlist<integer>();  Public voidSolveChar[] board) {        if(board = =NULL|| Board.length = = 0)            return; intm =board.length; intn = board[0].length; //merge O ' s on left & right boarder         for(inti = 0; I < m; i++) {            if(board[i][0] = = ' O ') {BFS (board, I,0); }             if(board[i][n-1] = = ' O ') {BFS (board, I, N-1); }        }         //Merge O ' s on top & bottom boarder         for(intj = 0; J < N; J + +) {            if(Board[0][j] = = ' O ') {BFS (board,0, J); }             if(Board[m-1][j] = = ' O ') {BFS (board, M-1, J); }        }         //Process the Board         for(inti = 0; I < m; i++) {             for(intj = 0; J < N; J + +) {                if(Board[i][j] = = ' O ') {Board[i][j]= ' X '; } Else if(Board[i][j] = = ' # ') {Board[i][j]= ' O '; }            }        }    }     Private voidBFsChar[] board,intIintj) {intn = board[0].length; //fill Current first and then its neighborsFillcell (board, I, J);  while(!Queue.isempty ()) {            intCur =Queue.poll (); intx = cur/N; inty = cur%N; Fillcell (board, X-1, y); Fillcell (board, X+ 1, y); Fillcell (board, X, y-1); Fillcell (board, X, y+ 1); }    }     Private voidFillcell (Char[] board,intIintj) {intm =board.length; intn = board[0].length; if(I < 0 | | I >= m | | J < 0 | | J >= N | | board[i][j]! = ' O ')            return; //Add current cell is the queue & then process it neighbors in BFSQueue.offer (i * n +j); BOARD[I][J]= ' # ';//use # to identify the O to keep    }}

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2) Dfs Depth first search

 Public voidSolveChar[] board) {    if(board = =NULL|| Board.length==0)         return; intm =board.length; intn = board[0].length; //merge O ' s on left & right boarder     for(inti=0;i<m;i++){        if(board[i][0] = = ' O ') {Merge (board, I,0); }         if(board[i][n-1] = = ' O ') {Merge (board, I,n-1); }    }     //Merge O ' s on top & bottom boarder     for(intj=0; j<n; J + +){         if(Board[0][j] = = ' O ') {Merge (board,0, J); }         if(Board[m-1][j] = = ' O ') {Merge (board, M-1, J); }    }     //Process the Board     for(inti=0;i<m;i++){         for(intj=0; j<n; J + +){            if(Board[i][j] = = ' O ') {Board[i][j]= ' X '; }Else if(Board[i][j] = = ' # ') {Board[i][j]= ' O '; }        }    }}  Public voidMergeChar[] board,intIintj) {    if(i<0 | | i>=board.length | | j<0 | | j>=board[0].length)return; if(board[i][j]! = ' O ')        return; BOARD[I][J]= ' # '; //recursive implementation of depth-first searchMerge (board, i-1, J); Merge (board, I+1, J); Merge (board, I, J-1); Merge (board, I, J+1);}

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Leetcode: Array