题目：原题链接（中等）
标签：字符串、回溯算法
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O(3N+4M) | O(3N+4M) | 32ms (95.50%) |
Ans 2 (Python) | O(3N+4M) | O(3N+4M) | 32ms (95.50%) |
Ans 3 (Python) |
LeetCode的Python执行用时随缘，只要时间复杂度没有明显差异，执行用时一般都在同一个量级，仅作参考意义。
解法一（枚举法）：
def letterCombinations(self, digits: str) -> List[str]:
phone = {"2": ["a", "b", "c"],
"3": ["d", "e", "f"],
"4": ["g", "h", "i"],
"5": ["j", "k", "l"],
"6": ["m", "n", "o"],
"7": ["p", "q", "r", "s"],
"8": ["t", "u", "v"],
"9": ["w", "x", "y", "z"]}
ans = [""]
for d in digits:
if d in phone:
new = []
for item in ans:
for c in phone[d]:
new.append(item + c)
ans = new
return ans解法二（不需要剪枝的回溯算法）：
测试用例中不包含字符1和字符0
def letterCombinations(self, digits: str) -> List[str]:
phone = {"2": ["a", "b", "c"],
"3": ["d", "e", "f"],
"4": ["g", "h", "i"],
"5": ["j", "k", "l"],
"6": ["m", "n", "o"],
"7": ["p", "q", "r", "s"],
"8": ["t", "u", "v"],
"9": ["w", "x", "y", "z"]}
def backtrack(now, next_digits):
if not next_digits:
ans.append(now)
else:
for d in phone[next_digits[0]]:
backtrack(now + d, next_digits[1:])
ans = []
if digits:
backtrack("", digits)
return ans
