【剑指offer】【双指针】 57-II.和为s的连续正数序列

题目链接：https://leetcode-cn.com/problems/he-wei-sde-lian-xu-zheng-shu-xu-lie-lcof/

双指针

[l,r]的区间和：s = (l + r) * (r - l + 1) / 2
通过利用l和r两个指针，初始l=1,r=2;
如果s == target，将[l,r]的数组添加到结果res中，l++；
如果s < target, r++;
如果s > target, l++;
时间复杂度O(n),空间复杂度O(1)

class Solution {
public:
    vector<vector<int>> findContinuousSequence(int target) {
        vector<vector<int>> res;
        vector<int> v;
        int l = 1, r = 2;
        while(l < r)
        {
            int sum = (l + r) * (r - l + 1) / 2;
            if(sum == target){
                v.clear();
                for(int i = l; i <= r; ++i) v.push_back(i);
                res.push_back(v);
                l++;
            }
            else if(sum < target) r++;
            else l++;
        }
        return res;
        
    }
};

滑动窗口

不使用求和公式，比双指针效率更高些，减少多余的计算；
时间复杂度O(n),空间复杂度O(1)

class Solution {
public:
    vector<vector<int>> findContinuousSequence(int target) {
        vector<vector<int>> res;
        for(int i = 1, j = 1, s = 1; i <= target / 2; i++)
        {
            while(s < target) j++, s += j;
            if(s == target && j > i)
            {
                vector<int> v;
                for(int k = i; k <= j; k++) v.push_back(k);
                res.push_back(v);
            }
            s -= i;
        }
        return res;
        
    }
};