LeetCode 之 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

.

要找固定长度的递增子序列，我们可以用两个元素存放满足关系的i,j的值，对于之后来的新元素如果大于arr[j]返回true，如果小于arr[i]更新i,如果在arr[i],arr[j]之间就更新j,遍历整个数组就行了，代码如下：

class Solution {
public:
bool increasingTriplet(vector<int>& nums) {
if(!nums.size()) return false;
int length=nums.size();
int min=0,se_min=-1;
for(int i=1;i<length;i++){
if(se_min==-1){
if(nums[i]>nums[min])
se_min=i;
if(nums[i]<nums[min])
min=i;
}else if(nums[i]>nums[se_min]){
return true;
}else if(nums[i]<=nums[min]){
min=i;
}else if(nums[i]<nums[se_min]){
se_min=i;
}
}
return false;
}
};