题目描述
小明很喜欢数学,有一天他在做数学作业时,要求计算出9~16的和,他马上就写出了正确答案是100。但是他并不满足于此,他在想究竟有多少种连续的正数序列的和为100(至少包括两个数)。没多久,他就得到另一组连续正数和为100的序列:18,19,20,21,22。现在把问题交给你,你能不能也很快的找出所有和为S的连续正数序列? Good Luck!
输出描述:
输出所有和为S的连续正数序列。序列内按照从小至大的顺序，序列间按照开始数字从小到大的顺序
解题思路
- 快慢指针的使用
- 首先要知道循环结束的条件，begin从开头 到 sum/2处，想一想，如果要加sum/2以后的数字，一定是要比sum大的
- 然后如果更新了begin，end，记得更新curSum
- 比如当curSum == sum时，begin++， end++， 此时要重新计算curSum
- 当curSum < sum的时候一样，end++后，curSum更新
- 同理 curSum > sum
class Solution {
public:
vector<vector<int>> findContinuousSequence(int target) {
vector<vector<int>> res;
int begin = 1;
int end = begin;
int sum = 0;
while(begin <= target / 2){
sum += end;
if(sum == target){
vector<int> v;
for(int i = begin; i <= end; i++){
v.push_back(i);
}
res.push_back(v);
sum = 0;
begin = begin + 1;
end = begin;
}else if(sum < target){
end++;
}else {
sum = 0;
begin = begin + 1;
end = begin;
}
}
return res;
}
};
class Solution {
public:
vector<vector<int> > FindContinuousSequence(int sum) {
vector<vector<int> > res;
int begin = 1;
int end = 2;
int mid = (sum + 1) / 2;
int curSum = begin + end;
while(begin <= mid){
if(curSum == sum){
vector<int> v;
for(int i = begin; i <= end; i++){
v.push_back(i);
}
res.push_back(v);
begin++;
end = begin + 1;
curSum = begin + end;
}else if(curSum < sum){
end++;
curSum += end;
}else{
curSum -= begin;
begin++;
}
}
return res;
}
};
class Solution {
public:
vector<vector<int> > FindContinuousSequence(int sum) {
vector<vector<int>> res;
int begin = 1;
int end = begin + 1;
int cur = 0;
while(begin <= sum / 2){
int n = end - begin + 1;
cur = n * (begin + end) / 2;
if(cur == sum){
vector<int> t;
for(int i = begin; i <= end; i++){
t.push_back(i);
}
res.push_back(t);
begin ++;
end = begin + 1;
}
else if(cur < sum){
end++;
}else{
begin++;
}
}
return res;
}
};
