Given a rectangular matrix of English lowercase letters board and a string word, your task is to find the number of occurrences of word in the rows(→), columns(↓) and diagonals(↘) of board.
For Example
eg1
board = [['s', 'o', 's', 'o'],
['s', 'o', 'o', 's'],
['s', 's', 's', 's']]
and word = "sos",
the output should be solution(board, word) = 3.
There are 2 occurrences of word starting from board[0][0](going → and ↘), and one starting from board[0][2](going ↓).
No other occurrences of word were counted, so the answer is 3.
eg2
board = [['a', 'a'],
['a', 'a']]
and word = "aa", the output should be
solution(board, word) = 5.
There are 2 horizontal, 2, vertical, and 1 diagonal occurrence of word, for a total of 5.
implement this function
func solution(board [][]string, word string) int {}
解法一：模拟法
func solution11(board [][]string, word string) int {
rows := len(board)
if rows == 0 {
return 0
}
cols := len(board[0])
res := 0
// 遍历矩阵的每一个元素
for i := 0; i < rows; i++ {
for j := 0; j < cols; j++ {
// 检查行方向
if j+len(word) <= cols {
match := true
for idx := 0; idx < len(word); idx++ {
if board[i][j+idx] != string(word[idx]) {
match = false
break
}
}
if match {
res++
}
}
// 检查列方向
if i+len(word) <= rows {
match := true
for idx := 0; idx < len(word); idx++ {
if board[i+idx][j] != string(word[idx]) {
match = false
break
}
}
if match {
res++
}
}
// 检查对角线方向
if i+len(word) <= rows && j+len(word) <= cols {
match := true
for idx := 0; idx < len(word); idx++ {
if board[i+idx][j+idx] != string(word[idx]) {
match = false
break
}
}
if match {
res++
}
}
}
}
return res
}
