leetcode(845) Longest Mountain in Array

寻找最长山脉子数组

最新推荐文章于 2021-08-27 15:10:38 发布

原创最新推荐文章于 2021-08-27 15:10:38 发布 · 697 阅读

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本文介绍了一种寻找数组中最长‘山脉’子数组的方法。‘山脉’定义为长度至少为3的子数组，并且存在一个转折点使得左右两侧分别递增和递减。通过状态机思想，文章详细阐述了如何高效地找到最长山脉子数组及其长度。

题：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.

题还是比较简单的。这个题我做的过程中，关键是要考虑全什么时候得计算result。要考虑全面。

int longestMountain(vector<int>& A) {

int result = 0;
if (A.size()<3)
return 0;
int left = 0;
int right = 0;
int state = 1;//state==1表示向上，state==-1表示向下
//考虑到result什么时候需要进行计算很重要
for (int i = 1; i<A.size(); i++)
{
if (A[i]>A[i - 1])
{
if (state == 1)
left++;
else
{
int tmp = right + left;
result = max(result, tmp);
left = 1;
right = 0;
state = 1;
}
}
else
{
if (A[i] == A[i - 1])
{
if (left > 0 && right > 0)
{
int tmp = right + left;
result = max(result, tmp);
}
left = 0;
}
else
if (left>0)
{
right++;
state =-1;
}
}
}
if (left>0 && right>0)
result = max(result, right + left);
return result + 1 >= 3 ? result + 1 : 0;
}