Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.
Formally the function should:
Return true if there exists i, j, k
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.
Your algorithm should run in O(n) time complexity and O(1) space complexity.
Examples:
Given [1, 2, 3, 4, 5]
,
return true
.
Given [5, 4, 3, 2, 1]
,
return false
.
用一个3个元素的数组记录3个递增子串。
当有新的元素比最大小时,替换到合适位置。
class Solution {
public:
bool increasingTriplet(vector<int>& nums) {
if(nums.size()<=3)
return false;
int Len = 1;
vector<int> Len_Min(4,0);
Len_Min[1] = nums[0];
for(int i=1; i<nums.size(); ++i){
if(nums[i]>Len_Min[Len]){
Len++;
if(Len>=3)
return true;
Len_Min[Len] = nums[i];
}
else{//untill nums[i]<Len_Min[k],replace the min one.
int k = Len;
while(k>=1 && nums[i]<=Len_Min[k]) k--;
Len_Min[k+1] = nums[i];
}
}
if(Len<3)
return false;
}
};