Question

334. 递增的三元子序列

English Version

题目描述

给你一个整数数组 nums ，判断这个数组中是否存在长度为 3 的递增子序列。

如果存在这样的三元组下标 (i, j, k) 且满足 i < j < k ，使得 nums[i] < nums[j] < nums[k] ，返回 true ；否则，返回 false 。

 

示例 1：

输入：nums = [1,2,3,4,5]
输出：true
解释：任何 i < j < k 的三元组都满足题意

示例 2：

输入：nums = [5,4,3,2,1]
输出：false
解释：不存在满足题意的三元组

示例 3：

输入：nums = [2,1,5,0,4,6]
输出：true
解释：三元组 (3, 4, 5) 满足题意，因为 nums[3] == 0 < nums[4] == 4 < nums[5] == 6

 

提示：

• 1 <= nums.length <= 5 * 105
• -231 <= nums[i] <= 231 - 1

 

进阶：你能实现时间复杂度为 O(n) ，空间复杂度为 O(1) 的解决方案吗？

解法

方法一

class Solution:
  def increasingTriplet(self, nums: List[int]) -> bool:
    mi, mid = inf, inf
    for num in nums:
      if num > mid:
        return True
      if num <= mi:
        mi = num
      else:
        mid = num
    return False

class Solution {
  public boolean increasingTriplet(int[] nums) {
    int n = nums.length;
    int[] lmi = new int[n];
    int[] rmx = new int[n];
    lmi[0] = Integer.MAX_VALUE;
    rmx[n - 1] = Integer.MIN_VALUE;
    for (int i = 1; i < n; ++i) {
      lmi[i] = Math.min(lmi[i - 1], nums[i - 1]);
    }
    for (int i = n - 2; i >= 0; --i) {
      rmx[i] = Math.max(rmx[i + 1], nums[i + 1]);
    }
    for (int i = 0; i < n; ++i) {
      if (lmi[i] < nums[i] && nums[i] < rmx[i]) {
        return true;
      }
    }
    return false;
  }
}

class Solution {
public:
  bool increasingTriplet(vector<int>& nums) {
    int mi = INT_MAX, mid = INT_MAX;
    for (int num : nums) {
      if (num > mid) return true;
      if (num <= mi)
        mi = num;
      else
        mid = num;
    }
    return false;
  }
};

func increasingTriplet(nums []int) bool {
  min, mid := math.MaxInt32, math.MaxInt32
  for _, num := range nums {
    if num > mid {
      return true
    }
    if num <= min {
      min = num
    } else {
      mid = num
    }
  }
  return false
}

function increasingTriplet(nums: number[]): boolean {
  let n = nums.length;
  if (n < 3) return false;
  let min = nums[0],
    mid = Number.MAX_SAFE_INTEGER;
  for (let num of nums) {
    if (num <= min) {
      min = num;
    } else if (num <= mid) {
      mid = num;
    } else if (num > mid) {
      return true;
    }
  }
  return false;
}

impl Solution {
  pub fn increasing_triplet(nums: Vec<i32>) -> bool {
    let n = nums.len();
    if n < 3 {
      return false;
    }
    let mut min = i32::MAX;
    let mut mid = i32::MAX;
    for num in nums.into_iter() {
      if num <= min {
        min = num;
      } else if num <= mid {
        mid = num;
      } else {
        return true;
      }
    }
    false
  }
}

方法二

class Solution {
  public boolean increasingTriplet(int[] nums) {
    int min = Integer.MAX_VALUE, mid = Integer.MAX_VALUE;
    for (int num : nums) {
      if (num > mid) {
        return true;
      }
      if (num <= min) {
        min = num;
      } else {
        mid = num;
      }
    }
    return false;
  }
}