LeetCode 334 Increasing Triplet Subsequence （两种方法）

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

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

题目分析：
法1：容易联想到nlogn求LIS的方法，这里只要判断是否存在长度为3的上升子序列，所以O1空间，On时间即可

class Solution {
public boolean increasingTriplet(int[] nums) {
int n = nums.length;
if (n < 3) {
return false;
}
int top = 0;
int[] stk = new int[5];
stk[++ top] = nums[0];
for (int i = 1; i < n; i++) {
if (nums[i] > stk[top]) {
stk[++ top] = nums[i];
} else {
for (int j = 1; j <= top; j++) {
if (stk[j] >= nums[i]) {
stk[j] = nums[i];
break;
}
}
}
if (top == 3) {
return true;
}
}
return false;
}
}

法2：直接维护当前最小和次小，如果某次不能更新，则说明存在

class Solution {
public boolean increasingTriplet(int[] nums) {
int ma = 2147483647, mi = 2147483647;
for (int i = 0; i < nums.length; i++) {
if (nums[i] <= mi) {
mi = nums[i];
} else if (nums[i] <= ma) {
ma = nums[i];
} else {
return true;
}
}
return false;
}
}