[LintCode/LeetCode] Search in Rotated Sorted Arra

U2FsdGVkX1x 發(fā)布于2019-08-14 15:41 / 3621人閱讀

摘要：找中點(diǎn)若起點(diǎn)小于中點(diǎn)，說明左半段沒有旋轉(zhuǎn)，否則說明右半段沒有旋轉(zhuǎn)。在左右半段分別進(jìn)行二分法的操作。只判斷有無，就容易了。還是用二分法優(yōu)化

Search in Rotated Sorted Array Problem

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Example

For [4, 5, 1, 2, 3] and target=1, return 2.

For [4, 5, 1, 2, 3] and target=0, return -1.

Challenge

O(logN) time

Note

找中點(diǎn)：若起點(diǎn)小于中點(diǎn)，說明左半段沒有旋轉(zhuǎn)，否則說明右半段沒有旋轉(zhuǎn)。
在左右半段分別進(jìn)行二分法的操作。

Solution

public class Solution {
    public int search(int[] A, int target) {
        int start = 0, end = A.length-1, mid = 0;
        while (start <= end) {
            mid = (start+end)/2;
            if (A[mid] == target) return mid;
            if (A[start] <= A[mid]) {
                if (A[start] <= target && target <= A[mid]) end = mid-1;
                else start = mid+1;
            }
            else {
                if (A[mid] < target && target <= A[end]) start = mid+1;
                else end = mid-1;
            }
        }
        return -1;
    }
}

Search in Rotated Sorted Array II Problem

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

Note

只判斷有無，就容易了。

Solution

還是用二分法優(yōu)化

public class Solution {
    public boolean search(int[] nums, int target) {
        if (nums == null || nums.length == 0) return false;
        int start = 0, end = nums.length-1;
        while (start <= end) {
            int mid = start+(end-start)/2;
            if (nums[mid] == target || nums[start] == target || nums[end] == target) return true;
            if (nums[start] < nums[mid]) {
                if (nums[start] <= target && target < nums[mid]) end = mid-1;
                else start = mid+1;
            }
            else if (nums[start] > nums[mid]) {
                if (nums[mid] < target && target <= nums[end]) start = mid+1;
                else end = mid-1;
            }
            else {
                if (nums[start] != target) start++;
                if (nums[end] != target) end--;
            }
        }
        return false;
    }
}

Updated 2018-08

class Solution {
    public boolean search(int[] nums, int target) {
        int start = 0, mid = 0, end = nums.length-1;
        while (start <= end) {
            mid = start+(end-start)/2;
            if (nums[mid] == target || nums[start] == target || nums[end] == target) return true;
            if (nums[start] < nums[mid]) {
                if (nums[start] < target && target < nums[mid]) end = mid-1;
                else start = mid+1;
            } else if (nums[start] > nums[mid]) {
                if (nums[mid] < target && target < nums[end]) start = mid+1;
                else end = mid-1;
            } else {
                start++;
                end--;
            }
        }
        return false;
    }
}