参考链接： https://leetcode-cn.com/problems/he-wei-sde-lian-xu-zheng-shu-xu-lie-lcof/solution/shi-yao-shi-hua-dong-chuang-kou-yi-ji-ru-he-yong-h/
滑动窗口
class
Solution {
public:
vector
<
vector
<
int
>>
findContinuousSequence(
int
target) {
//滑动窗口,L为左边界,R为右边界[L,R]
//L、R just swipe right,
vector
<
vector
<
int
>>
res;
//The meaning of the question can be seen from greater than or equal to1开始
int
L
=
1,
R
=
1;
int
sum
=
L;
//Because it is a continuous sequence of positive numbers,L最多到target/2 ,Because it contains at least two numbers,超过target/2 ,must be greater than target
while(
L
<=
target
/
2)
{
if(
sum
==
target)
//L往右滑动
{
vector
<
int
>
tmp;
for(
int
i
=
L;
i
<=
R;
i
++)
tmp.
push_back(
i);
res.
push_back(
tmp);
//Maintenance is also required after the value is found
sum
-=
L;
L
++;
}
else
if(
sum
<
target)
//R往右滑动,
{
R
++;
sum
+=
R;
}
else
if(
sum
>
target)
//L往右滑动,
{
sum
-=
L;
L
++;
}
}
return
res;
}
};
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滑动窗口+双端队列
class
Solution {
public:
vector
<
vector
<
int
>>
findContinuousSequence(
int
target) {
//滑动窗口,L为左边界,R为右边界[L,R]
//L、R just swipe right,
vector
<
vector
<
int
>>
res;
//The deque maintains a sliding window at this point
deque
<
int
>
record;
//The meaning of the question can be seen from greater than or equal to1开始
int
L
=
1,
R
=
1;
int
sum
=
L;
record.
push_back(
L);
//Because it is a continuous sequence of positive numbers,L最多到target/2 ,Because it contains at least two numbers,超过target/2 ,must be greater than target
while(
L
<=
target
/
2)
{
if(
sum
==
target)
//L往右滑动
{
vector
<
int
>
tmp(
record.
begin(),
record.
end());
res.
push_back(
tmp);
//Maintenance is also required after the value is found
sum
-=
L;
L
++;
record.
pop_front();
}
else
if(
sum
<
target)
//R往右滑动,
{
R
++;
record.
push_back(
R);
sum
+=
R;
}
else
if(
sum
>
target)
//L往右滑动,
{
sum
-=
L;
record.
pop_front();
L
++;
}
}
return
res;
}
};
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