資訊專欄INFORMATION COLUMN
摘要：當(dāng)遞歸到第次時，被調(diào)用了次。說明整個已經(jīng)被找到，返回?；氐胶瘮?shù)，遍歷整個數(shù)組，當(dāng)函數(shù)返回時，才返回否則在循環(huán)結(jié)束之后返回。
Problem
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.
ExampleGiven board =
[ "ABCE", "SFCS", "ADEE" ]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.
建立helper函數(shù)，當(dāng)board[i][j]和word的第一個字符相等，將board[i][j]置為非字母的其它字符，防止這個元素再一次被調(diào)用，然后遞歸調(diào)用helper函數(shù)判斷board[i][j]的上下左右相鄰的字符是否和word的下一個字符相等并將結(jié)果存入res，再把board[i][j]置回原來的字符（可能和word.charAt(0)相同的字符在board[][]中有多種情況，而此解并非正解，故還原數(shù)組在main函數(shù)里繼續(xù)循環(huán)），然后遞歸返回res到上一層helper函數(shù)。
當(dāng)遞歸到第word.length次時，helper被調(diào)用了word.length+1次。說明整個word已經(jīng)被找到，返回true。
回到main函數(shù)，遍歷整個board數(shù)組，當(dāng)helper函數(shù)返回true時，才返回true；否則在循環(huán)結(jié)束之后返回false。
class Solution {
public boolean exist(char[][] board, String word) {
if (board == null || board.length == 0 || board[0].length == 0) return word == null || word.length() == 0;
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (board[i][j] == word.charAt(0) && helper(board, i, j, 0, word)) {
return true;
}
}
}
return false;
}
private boolean helper(char[][] board, int i, int j, int index, String word) {
if (index == word.length()) return true;
if (i >= 0 && i < board.length &&
j >= 0 && j < board[0].length
&& board[i][j] == word.charAt(index)) {
board[i][j] = "#";
boolean res = helper(board, i+1, j, index+1, word) ||
helper(board, i-1, j, index+1, word) ||
helper(board, i, j+1, index+1, word) ||
helper(board, i, j-1, index+1, word);
board[i][j] = word.charAt(index);
return res;
}
return false;
}
}
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