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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

用 min, mid 记录遍历过程中遇到的最小值以及中间值,若出现 num > mid,说明找到了满足题目的三元组,返回 true;否则遍历结束返回 false。

Python3

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

Java

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;
    }
}

空间优化:

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;
    }
}

TypeScript

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;
}

C++

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;
    }
};

Go

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
}

Rust

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
    }
}

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