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.
Note: Your algorithm should run in O(n) time complexity and O(1) space complexity.
Example 1:
Input: [1,2,3,4,5]
Output: true
Example 2:
Input: [5,4,3,2,1]
Output: false
class Solution {
public:
//直接找出这个最长公共子串
//用一个数组保存最长公共子串，
//遍历原始数组，若nums[i] 小于最长公共子串数组开头数，则替换
//若nums[i]大于最长公共子串数组结尾数，则加入数组
//若nums[i]在最长公共子串数组中间位置，那么找到有序数组中第一个大于nums[i]的数，替换。
bool increasingTriplet(vector<int>& nums) {
if(nums.size()<3)
return false;
//存放最长递增子序列。
vector<int> dp;
dp.push_back(nums[0]);
for(int i=1;i<nums.size();i++){
if(nums[i] > dp.back()){
dp.push_back(nums[i]);
}else if(nums[i] < dp[0]){
dp[0] = nums[i];
}else{
//二分查找。查找所有大于key的元素中最左的元素。
int left = 0,right=dp.size()-1,target=nums[i];
while(left<=right){
int mid = left+(right-left)/2;
if(dp[mid] < target) left=mid+1;
else right=mid-1;
}
dp[left]=nums[i];
}
}
return dp.size() >= 3 ? true:false;
}
};
法二：
class Solution {
public:
//维护更新最小值和次小值。
bool increasingTriplet(vector<int>& nums) {
if(nums.size()<3) return false;
int smallest = INT_MAX;
int smaller = INT_MAX;
for(int i=0;i<nums.size();i++){
if(nums[i] <= smallest) smallest = nums[i];
else if(nums[i] <= smaller) smaller = nums[i];
else return true;
}
return false;
}
};