每日一題：山脈數組中查找目標值

示例1：
輸入：array = [1,2,3,4,5,3,1], target = 3
輸出：2
解釋：3 在數組中出現了兩次，下標分別爲 2 和 5，我們返回最小的下標 2。
示例2：
輸入：array = [0,1,2,4,2,1], target = 3
輸出：-1
解釋：3 在數組中沒有出現，返回 -1。

二分查找

因爲山脈數組由一個遞增序列和一個遞減序列兩個單調序列組成，所以先使用二分查找找到的峯值，然後再對兩個單調序列使用二分查找。就可以在O(logn)時間複雜度內找到目標值。

C++代碼

/**
 * // This is the MountainArray's API interface.
 * // You should not implement it, or speculate about its implementation
 * class MountainArray {
 *   public:
 *     int get(int index);
 *     int length();
 * };
 */

/*
先使用二分查找找到峯值
*/

class Solution {
    int binarySearch(MountainArray &mountainArr, int target, int left, int right, int key(int)) {
        target = key(target);
        while (left <= right) {
            int middle = (left + right) >> 1;
            int cur = key(mountainArr.get(middle));
            if (cur == target) return middle;
            else if (cur < target) left = middle + 1;
            else right = middle - 1;
        }
        return -1;
    }

public:
    int findInMountainArray(int target, MountainArray &mountainArr) {
        int left = 0, right = mountainArr.length() - 1; 
        // 使用二分查找找到峯值
        while (left < right) {
            int middle = (left + right) >> 1;
            if (mountainArr.get(middle) < mountainArr.get(middle + 1)) left = middle + 1;
            else right = middle;
        }

        int peak = left;
        int index = binarySearch(mountainArr, target, 0, peak, [](int x) -> int{return x;});
        if (index != -1) {
            return index;
        }
        return binarySearch(mountainArr, target, peak + 1, mountainArr.length() - 1, [](int x) -> int{return -x;});
    }
};

Python代碼

# """
# This is MountainArray's API interface.
# You should not implement it, or speculate about its implementation
# """
#class MountainArray:
#    def get(self, index: int) -> int:
#    def length(self) -> int:

class Solution:
    def binarySearch(self, target: int, mountain_arr: 'MountainArray', left: int, right: int, convert: 'function') -> int:
        target = convert(target)
        while left <= right:
            mid = (left + right) >> 1
            midEle = convert(mountain_arr.get(mid))
            if midEle == target:
                return mid
            elif midEle < target:
                left = mid + 1
            else:
                right = mid - 1
        return -1

    def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int:
        arrLen = mountain_arr.length()
        left = 0
        right = arrLen - 1
        while left < right:
            mid = (left + right) >> 1;
            if mountain_arr.get(mid) < mountain_arr.get(mid + 1):
                left = mid + 1
            else:
                right = mid
        
        peak = left
        index = self.binarySearch(target, mountain_arr, 0, peak, lambda x: x)
        if index != -1:
            return index
        return self.binarySearch(target, mountain_arr, peak + 1, arrLen - 1, lambda x: -x)