标签：第一个 rgs ret 范围 通过 sort main length ring
1、思路：先通过中间值和最后一个或者第一个元素比较，找出局部有序范围，再通过二分查找局部有序段
private static int sortArrFindOne(int arr[], int low, int high, int target) {
int mid = (high - low) / 2 + low;
if (arr[mid] == target) return mid;
if (arr[mid] < arr[high]) {
if (arr[mid] < target && target <= arr[high]) {
return find(arr, mid, high, target);
} else {
return sortArrFindOne(arr, low, mid, target);
}
} else {
if (arr[low] <= target && target < arr[mid]) {
return find(arr, low, mid, target);
} else {
return sortArrFindOne(arr, mid, high, target);
}
}
}
2、找局部有序（二分递归查找）
//局部有序
private static int find(int arr[], int low, int high, int target) {
int mid = (high - low) / 2 + low;
if (arr[mid] == target) return mid;
if (arr[mid] < target) {
return find(arr, mid + 1, high, target);
} else {
return find(arr, low, mid - 1, target);
}
}
3、测试
public static void main(String[] args) {
int arr[] = {6, 7, 8, 9, 1, 2, 3, 4, 5};
System.out.println(sortArrFindOne(arr, 0, arr.length - 1, 8));
System.out.println(sortArrFindOne(arr, 0, arr.length - 1, 1));
System.out.println(sortArrFindOne(arr, 0, arr.length - 1, 5));
System.out.println(sortArrFindOne(arr, 0, arr.length - 1, 6));
}
标签：第一个 rgs ret 范围 通过 sort main length ring
原文地址：https://www.cnblogs.com/junxing/p/11074837.html
