［LeetCode］Find Peak Element

解题思路：

比较数组中间两个数的大小，然后向数较大的那半边数组去寻找peak

前条件：left 指向区间最左边，right指向区间最右边的下一个；

不变式：mid ＝ (left + right) >> 1;

结束条件：mid == left，这说明，该区间，只有一个元素，也就是我们要找的peak

边界条件：如果硬要说边界条件，那么nums.size() == 0算是一个。

方法一：

时间复杂度为O(n)的方法

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        if (nums.size() <= 0) return -1;

        int peakIndex = 0;
        int i = 0;
        for (; i < nums.size(); ++i){
            if (i-1 < 0) continue;
            if (nums[i] < nums[i-1]) break;
        }

        peakIndex = i -1;
        return peakIndex;
    }
};

方法二：

时间复杂度为O(logn)的方法

二分查找的递归实现( 8ms )

思路：利用中间 mid 和 mid-1的相对大小，确定下一步查找区间范围

class Solution {
public:
    int findPeakElement(vector<int>& nums) {

        return binarySearch(nums, 0, nums.size());
    }

    int binarySearch(vector<int>&nums, int begin, int end){
        int mid = (begin + end) / 2;
        if ( mid == begin) return mid;

        int peak = 0;
        if (nums[mid] < nums[mid-1]){
            peak = binarySearch(nums, begin, mid);
        }else{
            peak = binarySearch(nums, mid, end);
        }

        return peak;
    }
};

二分查找的迭代实现：

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int left = 0;
        int right = nums.size();
        int mid = left;
        while(left < right){
            mid = (left + right) / 2;
            if (left == mid) break;
            if (nums[mid] > nums[mid-1]){
                left = mid;
            }else{
                right = mid;
            }
        }

        return left;
    }
};

二分查找的递归实现（盲目查找, 8 ms）反面教材，不推荐

思路：如果mid不是峰值，就向两边继续查找，不管mid两个数字的相对大小。

class Solution {
public:
    int findPeakElement(vector<int>& nums) {

        return binarySearch(nums, 0, nums.size());
    }

    int binarySearch(vector<int>&nums, int begin, int end){
        if (begin == end) return -1;

        int mid = (begin + end) / 2;

        if (((mid-1 < 0) || (mid-1>=0 && nums[mid] > nums[mid-1])) && 
            ((mid+1 >= nums.size()) || (mid+1 < nums.size() && nums[mid] > nums[mid+1]))){
                return mid;
            }

        int peak = -1;
        peak = binarySearch(nums, begin, mid);
        if (peak != -1) return peak;

        return binarySearch(nums, mid+1, end);
    }
};

来源：CSDN

作者：Lesley dude

链接：https://blog.csdn.net/aFeiOnePiece/article/details/47026623

标签

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