[leetcode][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?
解析:
找出最大的山形数组,这个数组必须是原数组的(连续的)子集,这个山形数组长度大于等于三,返回最长数组的长度。
参考答案(自己写的):
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class Solution {
public int longestMountain(int[] A) {
if (A.length < 3) return 0;
int result = 0;
int count = 0;
int top = 0;
for (int i = 1; i < A.length; i++) {
if (top == 0) {
if (A[i] > A[i - 1]) {
count++;
} else if (A[i - 1] > A[i] && count > 0) {
top = A[i - 1];
count++;
} else {
count=0;
}
} else {
if (A[i] < A[i - 1]) {
count++;
} else if (A[i - 1] < A[i - 2]) {
result = result > count + 1 ? result : count + 1;
count = 0;
top = 0;
}
if (A[i] > A[i-1]) {
count++;
}
}
}
if (top != 0 && count > 0) {
result = result > count + 1 ? result : count + 1;
}
return result;
}
}
遍历数组设置一个top来保存顶点,判断顶点(假设给定的数组都是大于0的正整数)是否为零来判断是“上山”还是“下山”。
参考答案(别人的):
public int longestMountain(int[] A) {
int res = 0, up = 0, down = 0;
for (int i = 1; i < A.length; ++i) {
if (down > 0 && A[i - 1] < A[i] || A[i - 1] == A[i]) up = down = 0;
if (A[i - 1] < A[i]) up++;
if (A[i - 1] > A[i]) down++;
if (up > 0 && down > 0 && up + down + 1 > res) res = up + down + 1;
}
return res;
}
别人的明显更简洁,效率也更高,少了很多判断,用up和down来分别计算“上山”和“下山”。