LeetCode 162. Find Peak Element https://leetcode.com/problems/find-peak-element/
题目描述
A peak element is an element that is strictly greater than its neighbors.
Given an integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.
You may imagine that nums[-1] = nums[n] = -∞.
- Example 1:
- Input: nums = [1,2,3,1]
- Output: 2
- Explanation: 3 is a peak element and your function should return the index number 2.
- Example 2:
- Input: nums = [1,2,1,3,5,6,4]
- Output: 5
- Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.
- Constraints:
- 1 <= nums.length <= 1000
- -231 <= nums[i] <= 231 - 1
- nums[i] != nums[i + 1] for all valid i.
解题思路
线性查找的极端情况是单峰且在另一端; 加入方向优化之后能缓解这种情况, 最差是单峰出现在最中间;
数据量再上去之和还可以考虑继续加入从中间到两端的查找, 四路并行查找 -- 优点类似于钻山隧道施工的意思.
另一种解决思路是 O(logN)的二分查找算法, 不是很直观能想到.
先找中间点, 如果恰好是一个峰, 当然可以直接返回;
如果不是峰, 则在较大的相邻元素一侧继续搜寻.
为什么是到大的一段去搜, 很简单的一个例子就能说明: 如 [1,2,3] 这种只有单峰的极端情况, 只有在较大侧才能找到峰.
参考代码
- /*
- * @lc App=leetcode id=162 lang=cpp
- *
- * [162] Find Peak Element
- */
- // @lc code=start
- class Solution {
- public:
- /*
- // nums[i] != nums[i + 1] for all valid i.
- int findPeakElement(vector<int>& nums) {
- if (nums.empty()) return -1;
- if (nums.size() == 1) return 0;
- int i = 0, j = nums.size() - 1;
- if (nums[i]> nums[i+1]) return i;
- if (nums[j-1] <nums[j]) return j;
- i++; j--;
- while (i <= j) {
- if (nums[i-1] < nums[i] && nums[i]> nums[i+1]) return i;
- if (nums[j-1] <nums[j] && nums[j]> nums[j+1]) return j;
- i++; j--;
- }
- assert(false);
- return -1;
- } // AC, direction-optimzed linear search, O(N)
- */
- // nums[i] != nums[i + 1] for all valid i.
- int findPeakElement(vector<int>& nums) {
- if (nums.empty()) return -1;
- if (nums.size() == 1) return 0;
- int l = 0, r = nums.size() - 1;
- while (l < r) {
- int m = l + (r - l) / 2;
- if (nums[m] < nums[m+1]) { // nums.size()==2 OK
- l = m + 1;
- } else {
- r = m;
- }
- }
- return l;
- } // AC, binary search, O(logN)
- };
- // @lc code=end
- [LeetCode 162.] Find Peak Element
来源: http://www.bubuko.com/infodetail-3719214.html