[LeetCode33]Search in Rotated Sorted Array

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#leetcode #数组 #二分查找

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本文介绍了一种在旋转排序数组中查找目标值的高效算法。通过分析数组旋转特性，利用二分搜索思想，实现O(log n)的时间复杂度。文章提供了Java和C++两种语言的实现代码。

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.

Analysis:

Binary Search

compare A[m]?target & A[m]?A[l]

Java

public int search(int[] A, int target) {
        int l = 0;
        int len = A.length;
        int r = len-1;
        while(l<=r){
        	int m = (l+r)/2;
        	if(A[m]==target) return m;
        	if(A[m]>=A[l]){
        		if(A[l]<=target && target<=A[m]) 
        			r = m-1;
        		else {
					l = m+1;
				}
        	}else {
				if(target>=A[m]&&target<=A[r])
					l = m+1;
				else {
					r = m-1;
				}
			}
        }
        return -1;
    }

c++

int search(int A[], int n, int target) {
        int l =0, r = n-1;
        while(l<=r){
            int m = (l+r)/2;
            if(A[m] == target) return m;
            if(A[m]>= A[l]){
                if(A[l]<=target && target<= A[m]) r = m-1;
                else l = m+1;
            }else{
                if(target>=A[l] || target<=A[m]) r = m-1;
                else l = m+1;
            }
        }
        return -1;
    }