81. Search in Rotated Sorted Array II

发表于 | 分类于

Suppose an array sorted in ascending order 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).

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

The array may contain duplicates.

解法1: O(logN), worst case O(n)

当有重复值的时候，我们在判断丢弃left或者是right half的时候，如果碰到nums[mid] == nums[end]或者nums[mid] == nums[start]的时候，不能判断哪一个部分需要抛弃。这个时候只能让end–或者start–
C++

1

Java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
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 + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                return true;
            } else if (nums[mid] < nums[end]) {
                if (nums[mid] < target && nums[end] >= target) {
                    start = mid;
                } else {
                    end = mid;
                }
            } else if (nums[mid] > nums[end]) {
                if (nums[start] <= target && nums[mid] > target) {
                    end = mid;
                } else {
                    start = mid;
                }
            } else {
                end--;  // can not remove a half
            }
        }
        if (nums[end] == target || nums[start] == target) {
            return true;
        } else {
            return false;
        }
    }
}