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.[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：
