A c++ solution using the lower_bound function(8ms).

  • 0
    class Solution {
    bool increasingTriplet(vector<int>& nums) {
        if(nums.size()<3) return false;
        for(int i=0;i<nums.size();i++)
            if(inc.size()>=3) return true;//important
            int ind=lower_bound(inc.begin(),inc.end(),nums[i])-inc.begin();
            else inc[ind]=nums[i];
        return inc.size()>=3;

    If without the condition that the size of the increasing sequence is 3, the complexity of this algorithm is O(nlogn). But since that the size is a constant number, and also the size of the vector 'inc' is a constant number, the time complexity is O(n) and the space is O(1).

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