【Leetcode】Increasing Triplet Subsequence

题目链接：https://leetcode.com/problems/increasing-triplet-subsequence/

题目：

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.

思路：

1、c[i] = max{c[j]+1,1} j<i 且nums[i]>nums[j]

时间复杂度O(n^2)，空间复杂度O(n)

2、

利用二分搜索，这里二分搜索的空间是常数项，所耗费的时间、空间都可以看成O(1)复杂度

算法1：

public boolean increasingTriplet(int[] nums) {      if (nums.length < 3)          return false;      int c[] = new int[nums.length];// c[i]表示从0~i 以nums[i]结尾的最长增长子串的长度      c[0] = 1;        for (int i = 1; i < nums.length; i++) {          int tmp = 1;          for (int j = 0; j < i; j++) {              if (nums[i] > nums[j]) {                  tmp = Math.max(c[j] + 1, tmp);              }          }          c[i] = tmp;          if (c[i] >= 3)              return true;      }      return false;  }  

算法2：

public boolean increasingTriplet(int[] nums) {      if (nums.length < 3)          return false;      int b[] = new int[3+1];// 长度为i的子串 最后一个数最小值      int end = 1;      b[end] = nums[0];        for (int i = 1; i < nums.length; i++) {          if (nums[i] > b[end]) {// 比最长子串最后元素还大，则更新最长子串长度              end++;              b[end] = nums[i];              if(end>=3)                  return true;          } else {// 否则更新b数组              int idx = binarySearch(b, nums[i], end);              b[idx] = nums[i];          }      }      return false;  }  /**  * 二分查找大于t的最小值，并返回其位置  */  public int binarySearch(int[] b, int target, int end) {      int low = 1, high = end;      while (low <= high) {          int mid = (low + high) / 2;          if (target > b[mid])              low = mid + 1;          else              high = mid - 1;      }      return low;  }  

算法2：