LeetCode 334. Increasing Triplet Subsequence（长度为3的递增子序列）

原题网址：https://leetcode.com/problems/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

.
方法：保存3个候选值，其中第1、2个位递增的一对，第3个位比第1个还小的数字。

public class Solution {
public boolean increasingTriplet(int[] nums) {
int count = 0;
int[] buf = new int[3];
for(int i=1; i<nums.length; i++) {
if (count == 0 && nums[i-1] < nums[i]) {
buf[0] = nums[i-1];
buf[1] = nums[i];
count = 2;
} else if (count == 2) {
if (buf[1] < nums[i]) return true;
if (buf[0] < nums[i]) {
buf[1] = nums[i];
} else if (buf[0] > nums[i]) {
buf[2] = nums[i];
count = 3;
}
} else if (count == 3) {
if (buf[1] < nums[i]) return true;
if (buf[2] >= nums[i]) {
buf[2] = nums[i];
} else {
buf[0] = buf[2];
buf[1] = nums[i];
count = 2;
}
}
}
return false;
}
}