给你一个 山脉数组 mountainArr,请你返回能够使得 mountainArr.get(index) 等于 target 最小 的下标 index 值。 如果不存在这样的下标 index,就请返回 -1。 何为山脉数组?如果数组 A 是一个山脉数组的话,那它满足如下条件: 首先,A.length >= 3 其次,在 0 < i < A.length - 1 条件下,存在 i 使得: A[0] < A[1] < … A[i-1] < A[i] A[i] > A[i+1] > … > A[A.length - 1]
你将 不能直接访问该山脉数组,必须通过 MountainArray 接口来获取数据:
MountainArray.get(k) - 会返回数组中索引为k 的元素(下标从 0 开始)
MountainArray.length() - 会返回该数组的长度
示例 1:
输入:array = [1,2,3,4,5,3,1], target = 3
输出:2
解释:3 在数组中出现了两次,下标分别为 2 和 5,我们返回最小的下标 2。
示例 2:
输入:array = [0,1,2,4,2,1], target = 3
输出:-1
解释:3 在数组中没有出现,返回 -1。
/*** // 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();* };*/class Solution {public:int binary_search1(int target, MountainArray & mountainArr, int left, int right){while(left <= right){int mid = (left + right) >> 1;int cur = mountainArr.get(mid);if (target == cur)return mid;else if(target < cur)right = mid-1;elseleft = mid+1;}return -1;}int binary_search2(int target, MountainArray & mountainArr, int left, int right){while(left <= right){int mid = (left + right) >> 1;int cur = mountainArr.get(mid);if (target == cur)return mid;else if(target < cur)left = mid+1;elseright = mid-1;}return -1;}int findInMountainArray(int target, MountainArray &mountainArr) {// binary search for find peak idxint len = mountainArr.length();int left = 0, right = len - 1;while(left < right){int mid = (left + right) >> 1;if (mountainArr.get(mid) < mountainArr.get(mid+1))left = mid+1;elseright = mid-1;}int peak = left;int ans_idx = binary_search1(target, mountainArr, 0, peak);if (ans_idx >= 0)return ans_idx;return binary_search2(target, mountainArr, peak+1, len-1);}};
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