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;
else
left = 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;
else
right = mid-1;
}
return -1;
}
int findInMountainArray(int target, MountainArray &mountainArr) {
// binary search for find peak idx
int 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;
else
right = 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);
}
};
欢迎交流,批评指正!