Discription:
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.
Solution:
bool increasingTriplet(vector<int>& nums) { //借鉴LIS，看长度是否大于3，超时
int len = nums.size();
if (len < 3)
return false;
vector<int> dp(len, 1);
for (int i = 1; i < len; i++){
for (int j = i - 1; j >= 0; j--){
if (nums[i]>nums[j]){
dp[i] = max(dp[i], dp[j] + 1);
}
}
if (dp[i] >= 3)
return true;
}
return false;
}
bool increasingTriplet(vector<int>& nums) {
int c1 = INT_MAX, c2 = INT_MAX;
for (int x : nums) {
if (x <= c1)
c1 = x;
else if (x <= c2)
c2 = x;
else
return true;
}
return false;
}GitHub-Leetcode:
https://github.com/wenwu313/LeetCode
