【LeetCode & 劍指offer刷題】數組題17:Increasing Triplet Subsequence
阿新 • • 發佈:2019-01-05
nac special include test mona wrap fff 17. 3.3
/*
判斷是否存在增序排列的三元子序列(元素不用連續)
用兩個臨時變量存儲第一個數和第二個數,x當做第三個數
例1:
input = [1 4 2 5 3]
c1,c2初始化為最大值INT_MAX
c1 = 1
c2 = 4 -> c2 =2 -> c3 =5 ,return true
例2:
[2 4 1 5]
c1 = 2,c2 = 4,c1 = 1,c3 = 5,return true,
能走到c3說明存在,但是c1,c2, c3存的數並不一定是符合要求的(順序可能不對)
O(n),O(1)
類似問題:Longest Increasing Subsequence
*/
#include <climits>
class Solution
{
public:
bool increasingTriplet(vector<int>& a)
{
if(a.size()<3) return false;
int c1 = INT_MAX,c2 = INT_MAX;
for(int x:a)
{
if(x<=c1) c1 = x; //c1為掃描過程中遇到的最小數,是第一個數的候選
else if(x<=c2) c2 = x; //當x>c1時,x可能為c2或者c3
else return true; //c1<c2<x,說明這樣的三元子序列存在
}
return false;
}
};
/*錯誤:沒有考慮元素可以不連續
class Solution
{
public:
bool increasingTriplet(vector<int>& a)
{
if(a.size()<3) return false;
for(int i = 0; i<a.size()-2; i++) //從第一個數掃描至倒數第3個元素
{
if(a[i]<a[i+1] && a[i+1] < a[i+2]) return true;
}
return false;
}
};*/
【LeetCode & 劍指offer 刷題筆記】目錄(持續更新中...)
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 ≤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. Credits: Special thanks to @DjangoUnchained for adding this problem and creating all test cases.i < j < k ≤ n-1 else return false.
【LeetCode & 劍指offer刷題】數組題17:Increasing Triplet Subsequence