【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
Letter Combinations of a Phone Number
Given a string containing digits from
2-9
inclusive, return all possible letter combinations that the number could represent.
A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
Example:
Input:
"23"
Output:
["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.
//传统手机数字盘的按键产生的所有可能字母组合
//感觉和回溯法没有什么关系，就是穷举法
class
Solution
{
public
:
vector
<
string
>
letterCombinations
(
string digits
)
{
if
(
digits
.
empty
())
return
vector
<
string
>();
//异常情况处理
unordered_map
<
char
,
string
>
map
=
{
{
'2'
,
"abc"
},
{
'3'
,
"def"
},
{
'4'
,
"ghi"
},
{
'5'
,
"jkl"
},
{
'6'
,
"mno"
},
{
'7'
,
"pqrs"
},
{
'8'
,
"tuv"
},
{
'9'
,
"wxyz"
}
};
vector
<
string
>
result
;
result
.
push_back
(
""
);
// add a seed for the initial case,因为之后会遍历结果向量 push了一个空元素，result size变为1
for
(
int
i
=
0
;
i
<
digits
.
size
();
i
++)
//遍历数字, 如 2 3
{
if
(
map
.
find
(
digits
[
i
])
==
map
.
end
())
continue
;
//如果不是2~9的数字，继续循环
string
&
candidate
=
map
[
digits
[
i
]];
//得到候选字符构成的字符串
vector
<
string
>
temp
;
for
(
string
&
ri
:
result
)
//遍历结果向量，如 "a" "b" "c"
{
for
(
char
ci
:
candidate
)
//遍历候选字符 如 "def"
{
temp
.
push_back
(
ri
+
ci
);
}
}
// result = temp; //复制到结果向量
result
.
swap
(
temp
);
//交换，swap does not take memory copy
}
return
result
;
}
};