• Leetcode: Increasing Triplet Subsequence


    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.

    Naive Solution: use DP, Time O(N^2), Space O(N)

      dp[i] represents the length of longest increasing subsequence till i including element i in nums array. dp[i] is initialized to be 1.

      dp[i] = max(dp[i], dp[j]+1), where j is an index before i

     1 public class Solution {
     2     public boolean increasingTriplet(int[] nums) {
     3         int[] dp = new int[nums.length];
     4         for (int i=0; i<nums.length; i++) {
     5             dp[i] = 1;
     6             for (int j=0; j<i; j++) {
     7                 if (nums[j] < nums[i]) {
     8                     dp[i] = Math.max(dp[i], dp[j]+1);
     9                 }
    10                 if (dp[i] == 3) return true;
    11             }
    12         }
    13         return false;
    14     }
    15 }

    Better Solution: keep two values. Once find a number bigger than both, while both values have been updated, return true.

    small: is the minimum value ever seen untill now

    big: the smallest value that has something before it that is even smaller. That 'something before it that is even smaller' does not have to be the current min value.

    Example:
    3,2,1,4,0,5

    When you see 5, min value is 0, and the smallest second value is 4, which is not after the current min value.

     1 public class Solution {
     2     public boolean increasingTriplet(int[] nums) {
     3         int small = Integer.MAX_VALUE, big = Integer.MAX_VALUE;
     4         for (int n : nums) {
     5             if (n <= small) {
     6                 small = n;
     7             }
     8             else if (n <= big) {
     9                 big = n;
    10             }
    11             else return true;
    12         }
    13         return false;
    14     }
    15 }
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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/6092267.html
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