- Find Peak Element
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] = -∞.
You must write an algorithm that runs in O(log n) time.
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.
class Solution {
public:
int binarySearch(int left,int right,vector<int>& nums){
int mid=(left+right)/2;
if(nums[mid]>nums[mid+1]&&nums[mid]>nums[mid-1]){
return mid;
}
if(nums[mid]<nums[mid+1]){
return binarySearch(mid,right,nums);
}
else{
return binarySearch(left,mid+1,nums);
}
}
int findPeakElement(vector<int>& nums) {
int right=nums.size()-1,left=0;
int mid=(right+left)/2;
if(nums.size()==1){
return 0;
}
else if(nums[0]>nums[1]){
return 0;
}
else if(nums[nums.size()-1]>nums[nums.size()-2]){
return nums.size()-1;
}
else{
while(!(nums[mid]>nums[mid+1]&&nums[mid]>nums[mid-1])){
if(nums[mid]<nums[mid+1]){
left=mid;
}
else{
right=mid+1;
}
mid=(left+right)/2;
}
return mid;
}
}
};
- 一开始读错题了,以为输出的是元素,实际上要输出位置
- 之前写了前面的函数,内存占用太高了,改用while循环了
想知道该怎样不每次都输入整个vector时也能用递归
题解三种方法,在这里,前两种都是O(n)时间复杂度,第三种二分法和我的应该差不多,没细看,有一些没用过的操作需要看看