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剑指 Offer 57 - II. 和为s的连续正数序列

思路

方法一：枚举 + 暴力

 

 1 class Solution {
 2 public:
 3     vector<vector<int>> findContinuousSequence(int target) {
 4         vector<vector<int>> res;
 5         for(int i = 1; i <= target/2; ++i) {
 6             vector<int> tmp;
 7             int sum = 0;
 8             for(int j = i; j < target && sum < target; ++j) {
 9                 sum += j;
10                 tmp.push_back(j);
11             }
12 
13             if(sum == target) {
14                 res.push_back(tmp);
15             }
16         }
17 
18         return res;
19     }
20 };

 

 

 1 class Solution {
 2 public:
 3     vector<vector<int>> findContinuousSequence(int target) {
 4         vector<vector<int>> vec;
 5         vector<int> res;
 6         int sum = 0, limit = (target - 1) / 2; // (target - 1) / 2 等效于 target / 2 下取整
 7         for (int x = 1; x <= limit; ++x) {
 8             long long delta = 1 - 4 * (x - 1ll * x * x - 2 * target);
 9             if (delta < 0) {
10                 continue;
11             }
12             int delta_sqrt = (int)sqrt(delta + 0.5);
13             if (1ll * delta_sqrt * delta_sqrt == delta && (delta_sqrt - 1) % 2 == 0) {
14                 int y = (-1 + delta_sqrt) / 2; // 另一个解(-1-delta_sqrt)/2必然小于0，不用考虑
15                 if (x < y) {
16                     res.clear();
17                     for (int i = x; i <= y; ++i) {
18                         res.push_back(i);
19                     }
20                     vec.push_back(res);
21                 }
22             }
23         }
24         return vec;
25     }
26 };

 

 

 1 class Solution {
 2 public:
 3     vector<vector<int>> findContinuousSequence(int target) {
 4         vector<vector<int>> res;
 5         int l = 1, r = 2;
 6         while(l < r) {
 7             int sum = (r-l+1)*(l+r)/2;
 8             if(sum == target) {
 9                 vector<int> tmp;
10                 for(int i = l; i <= r; ++i)
11                     tmp.push_back(i);
12                 res.push_back(tmp);
13                 ++l;
14             } else if(sum < target) {
15                 ++r;
16             } else {
17                 ++l;
18             }
19         }
20 
21         return res;
22     }
23 };

 

也可以不使用求和公式求sum，而是每次对指针进行移动的时候，同时对sum的值进行改变。代码如下：

 1 class Solution {
 2 public:
 3     vector<vector<int>> findContinuousSequence(int target) {
 4         vector<vector<int>> res;
 5         int i = 1, j = 2;
 6         int sum = 3;
 7         while(i < j) {
 8             if(sum < target) {
 9                 ++j;
10                 sum += j;
11             } else if (sum > target) {
12                 sum -= i;
13                 ++i;
14             } else {
15                 vector<int> v;
16                 for(int k = i; k <= j; ++k) {
17                     v.push_back(k);
18                 }
19                 res.push_back(v);
20                 sum -= i;
21                 ++i;
22             }
23         }
24 
25         return res;
26     }
27 };

 

参考

力扣官方题解 - 和为s的连续正数序列

posted @ 2020-11-14 09:41  拾月凄辰  阅读(106)  评论(0)  编辑  收藏  举报