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31 lines
1.2 KiB
Markdown
31 lines
1.2 KiB
Markdown
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# [162. Find Peak Element](https://leetcode.com/problems/find-peak-element/)
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# 思路
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寻找一个数组中的极大值. 由于要求对数复杂度, 所以我们很容易想到二分法.
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由于题目说了nums[-1]和nums[n]为负无穷, 那么极大值一定存在(想象一下被山谷包围的山脉).
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那么在二分的循环中:
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1. 若`mid == 0 || nums[mid-1] < nums[mid]`, 即mid左边是上升趋势, 那么mid或者其右边一定存在极大值;
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2. 满足1且`mid == n - 1 || nums[mid] > nums[mid+1]`, 即mid就是极大值;
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3. 满足1但不满足2, 那么其右边一定存在极大值, 此时更新`low = mid + 1`;
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4. 否则, mid左边一定存在极大值, 此时更新`high = mid - 1`;
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# C++
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``` C++
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class Solution {
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public:
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int findPeakElement(vector<int>& nums) {
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int n = nums.size();
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int low = 0, high = n - 1, mid;
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while(low < high){
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mid = low + (high - low) / 2;
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if(mid == 0 || nums[mid-1] < nums[mid]){
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if(mid == n - 1 || nums[mid] > nums[mid+1]) return mid; // 找到
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else low = mid + 1;
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}
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else high = mid - 1;
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}
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return low;
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}
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};
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```
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