Description
Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.
Example 1:
Input: nums = [1,2,3,4,5] Output: true Explanation: Any triplet where i < j < k is valid.
Example 2:
Input: nums = [5,4,3,2,1] Output: false Explanation: No triplet exists.
Example 3:
Input: nums = [2,1,5,0,4,6] Output: true Explanation: The triplet (3, 4, 5) is valid because nums[3] 0 < nums[4] 4 < nums[5] == 6.
Constraints:
1 <= nums.length <= 5 * 105-231 <= nums[i] <= 231 - 1
Follow up: Could you implement a solution that runs in O(n) time complexity and O(1) space complexity?
Code
利用 Longest Increasing Subsequence 去找到最長的 increasing subsequence，若長度超過三，代表符合題意的解答存在。
Time & Space
class Solution {
public:
bool increasingTriplet(vector<int>& nums) {
vector<int> res;
for(int i = 0; i < nums.size(); i++) {
if(res.empty() || res.back() < nums[i]) {
res.emplace_back(nums[i]);
} else {
auto it = lower_bound(res.begin(), res.end(), nums[i]);
*it = nums[i];
}
}
return res.size() >= 3;
}
};class Solution {
public:
bool increasingTriplet(vector<int>& nums) {
vector<int> LIS;
for(auto n: nums) {
if(LIS.empty()) LIS.push_back(n);
else {
int l = 0, r = LIS.size();
while(l < r) {
int m = (l + r) / 2;
if(LIS[m] >= n) {
r = m;
} else {
l = m + 1;
}
}
if(l == LIS.size()) LIS.push_back(n);
else {
LIS[l] = n;
}
}
}
return LIS.size() >= 3;
}
};two pointer
class Solution {
public:
bool increasingTriplet(vector<int>& nums) {
int c1 = INT_MAX, c2 = INT_MAX;
for (int x : nums) {
if (x <= c1) {
c1 = x; // c1 is min seen so far (it's a candidate for 1st element)
} else if (x <= c2) { // here when x > c1, i.e. x might be either c2 or c3
c2 = x; // x is better than the current c2, store it
} else { // here when we have/had c1 < c2 already and x > c2
return true; // the increasing subsequence of 3 elements exists
}
}
return false;
}
};