面试题57 - II. 和为s的连续正数序列¶
Tags: Easy
Math
Two Pointers
Links: https://leetcode-cn.com/problems/he-wei-sde-lian-xu-zheng-shu-xu-lie-lcof/
输入一个正整数 target
,输出所有和为 target
的连续正整数序列(至少含有两个数)。
序列内的数字由小到大排列,不同序列按照首个数字从小到大排列。
示例 1:
输入:target = 9
输出:[[2,3,4],[4,5]]
示例 2:
输入:target = 15
输出:[[1,2,3,4,5],[4,5,6],[7,8]]
限制:
1 <= target <= 10^5
class Solution {
public:
vector<vector<int>> findContinuousSequence(int target) {
std::ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long t = target;
vector<vector<int>> res;
long long mid = t / 2;
for (long long m = 1; m <= mid; ++m) {
int k = ((-2 * m - 1) + sqrt(4 * m * m - 4 * m + 8 * t + 1)) / 2;
if ((2 * m + k) * (k + 1) / 2 == t) {
vector<int> tmp(k + 1);
for (int i = 0; i <= k; ++i) {
tmp[i] = m + i;
}
res.push_back(tmp);
}
}
return res;
}
};
数学解法,等差数列求和,判断末尾项,检验总和是否相等,相等之后可以直接判断出序列长度,开一个数组,填入数字即可,注意只需要判断一半的数据即可。
双指针解法:
class Solution {
public:
vector<vector<int>> findContinuousSequence(int target) {
std::ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
vector<vector<int>> res;
long long start = 1, end = 1;
long long mid = target / 2;
while (start <= mid) {
long long sum = (start + end) * (end - start + 1) / 2;
if (sum == target) {
int k = end - start + 1;
vector<int> tmp(k);
for (int i = 0; i < k; ++i) {
tmp[i] = start + i;
}
res.push_back(tmp);
++start; ++end;
}
else if (sum < target) ++end;
else ++start;
}
return res;
}
};