LintCode--最长上升连续子序列

lintcode--longest-increasing-continuous-subsequence(最长上升连续子序列)

原题链接：http://www.lintcode.com/zh-cn/problem/longest-increasing-continuous-subsequence/

给定一个整数数组（下标从 0 到 n-1， n 表示整个数组的规模），请找出该数组中的最长上升连续子序列。（最长上升连续子序列可以定义为从右到左或从左到右的序列。）

样例

给定 [5, 4, 2, 1, 3], 其最长上升连续子序列（LICS）为 [5, 4, 2, 1], 返回 4.

给定 [5, 1, 2, 3, 4], 其最长上升连续子序列（LICS）为 [1, 2, 3, 4], 返回 4.

注意

time

分析：

从头遍历数组，分别用up和down记录最大连续上升序列和最大连续下降序列。

时间复杂度：O（n）     空间复杂度O（1）

代码（C++、Java、Python）：

class Solution {
public:
    /**
     * @param A an array of Integer
     * @return  an integer
     */
    int max(int a, int b){
        if (a > b) return a;
        else return b;
    }
    int longestIncreasingContinuousSubsequence(vector<int>& A) {
        // Write your code here
        int res = 0, up = 1, down = 1;
        int n = A.size();
        if (n <= 2) return n;
        for (int i = 1; i < n; i++){
            if (A[i] > A[i-1]){
                up += 1;
                down = 1;
            }
            else if (A[i] == A[i-1]){
                up += 1;
                down += 1;
            }
            else if (A[i] < A[i-1]){
                up = 1;
                down += 1;
            }
            if (up > res || down > res)
                res = max(up, down);
        }
        return res;
    }
};

class Solution {
public:
    /**
     * @param A an array of Integer
     * @return  an integer
     */
    int max(int a, int b){
        if (a > b) return a;
        else return b;
    }
    int longestIncreasingContinuousSubsequence(vector<int>& A) {
        // Write your code here
        int res = 0, up = 1, down = 1;
        int n = A.size();
        if (n <= 2) return n;
        for (int i = 1; i < n; i++){
            if (A[i] > A[i-1]){
                up += 1;
                down = 1;
            }
            else if (A[i] == A[i-1]){
                up += 1;
                down += 1;
            }
            else if (A[i] < A[i-1]){
                up = 1;
                down += 1;
            }
            if (up > res || down > res)
                res = max(up, down);
        }
        return res;
    }
};

class Solution:
    # @param {int[]} A an array of Integer
    # @return {int}  an integer
    def longestIncreasingContinuousSubsequence(self, A):
        # Write your code here
        res = 0
        up = 1
        down = 1
        n = len(A)
        if n <= 2:
            return n
        for i in range(1, n):
            if A[i] > A[i-1]:
                up += 1
                down = 1
            elif A[i] == A[i-1]:
                up += 1
                down += 1
            elif A[i] < A[i-1]:
                up = 1
                down += 1
            if up > res or down > res:
                res = max(up, down)
        return res
        
def max(a, b):
    if a > b:
        return a
    else:
        return b