(这是一个 交互式问题 )
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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/find-in-mountain-array
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
class Solution {
private MountainArray mountainArray;
private boolean isHigher(int x, int y) {
int n = mountainArray.length();
if (y == -1 || y == n) {
return true;
}
return mountainArray.get(x) > mountainArray.get(y);
}
private int findHighest() {
int left = 0, right = mountainArray.length() - 1;
while (left <= right) {
int mid = (left + right) >> 1;
if (isHigher(mid, mid + 1) && isHigher(mid, mid - 1)) {
return mid;
}
if (isHigher(mid + 1, mid)) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return -1;
}
private int search(int left, int right, int target, boolean asc) {
while (left <= right) {
int mid = (left + right) >> 1;
int midVal = mountainArray.get(mid);
if (midVal == target) {
return mid;
}
if (midVal < target) {
if (asc) {
left = mid + 1;
} else {
right = mid - 1;
}
} else {
if (asc) {
right = mid - 1;
} else {
left = mid + 1;
}
}
}
return -1;
}
public int findInMountainArray(int target, MountainArray mountainArr) {
this.mountainArray = mountainArr;
int highestIndex = findHighest();
int idx = search(0, highestIndex, target, true);
if (idx != -1) {
return idx;
}
idx = search(highestIndex, mountainArray.length() - 1, target, false);
return idx;
}
}
interface MountainArray {
default int get(int index) {
return 0;
}
default int length() {
return 0;
}
}