334. Increasing Triplet subsequence

I did it once, but I can't remember.

1Now set an X, record the smallest of the three numbers, Y record the middle one, initialize X is nums[0],y is Integer.max_value .2 3 then start looking back from 1:4 51. If this number is larger than Y, because this time x <y, if you find a third that is larger than Y, then return true6 72If this number is not larger than Y, there are two things:8 91) is smaller than x (or equal to x), the x is updated to the current position, but Y does not move, X is the minimum number of potentially next set, Y still indicates that the second number, if the next group found greater than Y, can still return true. Ten  One2no smaller than x, that is, between the current X and Y, then the update y,x unchanged, indicating that a group of XY failed. Why this must be found, because the X is already updated first (because the X is updated in 2.1), less than the original is less than X, Y (because the condition 2 illustrates this), so this is a smaller set of XY, more likely to succeed.  A  -If the end of the entire array of traversal has not found the answer, then return false

That's probably it. Hope to remember!!

1      Public BooleanIncreasingtriplet (int[] nums) {2         if(Nums.length < 3) {3             return false;4         }5         intx = nums[0];6         inty =Integer.max_value;7          for(inti = 1; i < nums.length; i++) {8             if(Nums[i] >y) {9                 return true;Ten}Else { One                 if(Nums[i] <=x) { Ax =Nums[i]; -}Else { -y =Nums[i]; the                 } -             } -         } -         return false; +}

Time complexity is O (n)

334. Increasing Triplet subsequence