leetcode-mid-backtracking -22. Generate Parentheses-79 Word Search -NO

mycode   错误,因为借鉴了Number of Islands问题中的方法,导致在for循环中即使已经出现了答案,也还会继续遍历。但是两个题目的不同时,island需要找出所有的情况,这个题只需要找到第一个完整结果就可以返回

 参考:

 
def exist(board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        def find(board, word, i, j, m, n):
            if word == '':
                print('if',i,j)
                return True
            if i < 0 or i >= m or j < 0 or j >= n:
                print('if2',i,j)
                return False
            elif word[0] == board[i][j]:
                print('elif',i,j)
                board[i][j] = None #标记
                res = find(board, word[1:], i+1, j, m, n)or find(board, word[1:], i-1, j, m, n)or find(board, word[1:], i, j+1, m, n)or find(board, word[1:], i, j-1, m, n)
                board[i][j] = word[0] #恢复原来的值
                return res
            print(i,j)
        if len(word) == 0:
            return True
        m = len(board)
        if m == 0:
            return False
        n = len(board[0])
        for i in range(m):
            for j in range(n):
                if find(board, word, i, j, m, n):
                    return True
        return False

 

 
上一篇:20. Valid Parentheses[E]有效的括号


下一篇:Leetcode 22 - Generate Parentheses.