Search in Rotated Sorted Array

lc 33

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.

步骤

1. 判断 start 与mid
2. 情况一，target 在不在start的mid 之间，
3. 情况二，在不在mid与end之间

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

        if (nums[start] == target) return start;
        else if ( nums[end] == target) return end;
        return -1;
    }
}

如果存在重复

public class Solution {
    public boolean search(int[] A, int target) {

        if (A == null || A.length == 0) {
            return false;
        }

        int start = 0;
        int end = A.length - 1;
        int mid;

        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (A[mid] == target) {
                return true;
            }
            if(A[start] == A[mid]){
                start++;
                continue;
            }
            if (A[start] < A[mid]) {
                // situation 1, red line
                if (A[start] <= target && target <= A[mid]) {
                    end = mid;
                } else {
                    start = mid;
                }
            } else {
                // situation 2, green line
                if (A[mid] <= target && target <= A[end]) {
                    start = mid;
                } else {
                    end = mid;
                }
            }
        } // while

        if (A[start] == target) {
            return true;
        }
        if (A[end] == target) {
            return true;
        }
        return false;
    }   

}