Search in Rotated Sorted Array, Leetcode 解题笔记

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.

Binary Search的变形,最优的思路如下:
Picture123

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

注意代码中判断大小时等号的位置。

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