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.
For example,
Given board =
[ ["ABCE"], ["SFCS"], ["ADEE"] ]word =
"ABCCED", -> returns true,"SEE", -> returns true,"ABCB", -> returns false.
典型的DFS题目。Coding过程中注意状态的变化以及边界情况。
class Solution {
public:
bool exist(vector<vector<char> > &board, string word) {
if (word == "") {
return false;
}
int nNumberOfRows = board.size();
vector<vector<bool> > bVisit;
for (int i = 0; i < nNumberOfRows; ++i) {
int nNumberOfCols = board[i].size();
bVisit.push_back(vector<bool>());
for (int j = 0; j < nNumberOfCols; ++j) {
bVisit[i].push_back(false);
}
}
for (int i = 0; i < nNumberOfRows; ++i) {
int nNumberOfCols = board[i].size();
for (int j = 0; j < nNumberOfCols; ++j) {
if (board[i][j] == word[0]) {
if (dfs(board, i, j, bVisit, word, 1)) {
return true;
}
}
}
}
return false;
}
protected:
bool dfs(vector<vector<char> > &board, int row, int col, vector<vector<bool> > &bVisit, string &word, int n) {
if (n == word.size()) {
return true;
}
bVisit[row][col] = true;
static int dir[][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
for (int i = 0; i < 4; ++i) {
int newRow = row + dir[i][0];
int newCol = col + dir[i][1];
if (isValid(board, bVisit, newRow, newCol) && board[newRow][newCol] == word[n]) {
if (dfs(board, newRow, newCol, bVisit, word, n + 1)) {
return true;
}
}
}
bVisit[row][col] = false;
return false;
}
bool isValid(vector<vector<char> > &board, vector<vector<bool> > &bVisit, int row, int col) {
if (row >= 0 && row < board.size() &&
col >= 0 && col < board[row].size() &&
!bVisit[row][col]) {
return true;
}
return false;
}
};leetcode-Word Search,布布扣,bubuko.com
原文：http://blog.csdn.net/freeliao/article/details/38093735