LeetCode 845. Longest Mountain in Array C++

最新推荐文章于 2021-10-01 11:07:13 发布

WX_ming

最新推荐文章于 2021-10-01 11:07:13 发布

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本文链接：https://blog.youkuaiyun.com/WX_ming/article/details/81557952

本文介绍了一种寻找数组中最长'山脉'子数组的方法。'山脉'定义为包含一个最大值，两侧分别递增和递减的部分，且不允许有相等元素。文章详细解释了如何通过一次遍历实现该目标，并保持O(1)的空间复杂度。

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845. Longest Mountain in Array

Let’s call any (contiguous) subarray B (of A) a mountain if the following properties hold:

B.length >= 3
There exists some 0 < i < B.length - 1 such that B[0] < B[1] < … B[i-1] < B[i] > B[i+1] > … > B[B.length - 1]
(Note that B could be any subarray of A, including the entire array A.)

Given an array A of integers, return the length of the longest mountain.

Return 0 if there is no mountain.

Example 1:

Input: [2,1,4,7,3,2,5]
Output: 5
Explanation: The largest mountain is [1,4,7,3,2] which has length 5.

Example 2:

Input: [2,2,2]
Output: 0
Explanation: There is no mountain.

Note:

0 <= A.length <= 10000
0 <= A[i] <= 10000

Follow up:

Can you solve it using only one pass?
Can you solve it in O(1) space?

Approach

给你一数组，问你数组中‘mountian’最长是多少，‘mountain’的定义是这一段数组中有一个最大值，以最大值为分界线左边是递增，右边是递减，当中不允许相等的元素，而且长度不小于3。这道题我的想法比较奇特，用到比较多的if-else ，首先做为mountian需要有一个上坡，我一开是我先要找到上坡，记录长度从一开始，如果当中有元素相等，那么重新记录长度从一开始，直至遇到下坡，当到达下坡的时候我就开始不断更新我的maxn（mountian的长度最大值），直至遇到上坡，我又重一开始记录长度，如果当中有元素相等，那么长度归为一，而且等待上坡再开始记录长度，时间复杂度为O(n),空间复杂度为O(1)。12ms

Code

class Solution {
public:
    int longestMountain(vector<int>& A) {
        if (A.size() < 3)return 0;
        bool up = true;
        int maxn = 0, cnt = 1;
        for (int i = 0; i < A.size()-1; i++) {
            if (up) {
                if (A[i] < A[i + 1]) {
                    cnt++;
                }
                else if(A[i]==A[i+1]){
                    cnt = 1;
                }
                else if(cnt>1&&A[i]>A[i+1]){
                    cnt++;
                    maxn = max(cnt, maxn);
                    up = false;
                }
            }
            else {
                if (A[i] > A[i + 1]&&cnt>1) {
                    cnt++;
                    maxn = max(cnt, maxn);
                }
                else if(A[i]==A[i+1]) {
                    cnt = 1;
                }
                else if(A[i]<A[i+1]){
                    cnt = 1;
                    cnt++;
                    up = true;
                }
            }
        }
        return maxn;
    }
};