Find Minimum in Rotated Sorted Array II详解编程语言

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).

Find the minimum element.

The array may contain duplicates.

Example

Given [4,4,5,6,7,0,1,2] return 0

题解

由于此题输入可能有重复元素，因此在num[mid] == num[end]时无法使用二分的方法缩小start或者end的取值范围。此时只能使用递增start/递减end逐步缩小范围。

C++

class Solution { 
public: 
    /** 
     * @param num: a rotated sorted array 
     * @return: the minimum number in the array 
     */ 
    int findMin(vector<int> &num) { 
        if (num.empty()) { 
            return -1; 
        } 
 
        vector<int>::size_type start = 0; 
        vector<int>::size_type end = num.size() - 1; 
        vector<int>::size_type mid; 
        while (start + 1 < end) { 
            mid = start + (end - start) / 2; 
            if (num[mid] > num[end]) { 
                start = mid; 
            } else if (num[mid] < num[end]) { 
                end = mid; 
            } else { 
                --end; 
            } 
        } 
 
        if (num[start] < num[end]) { 
            return num[start]; 
        } else { 
            return num[end]; 
        } 
    } 
};

Java

public class Solution { 
    /** 
     * @param num: a rotated sorted array 
     * @return: the minimum number in the array 
     */ 
    public int findMin(int[] num) { 
        if (num == null || num.length == 0) return Integer.MIN_VALUE; 
 
        int lb = 0, ub = num.length - 1; 
        // case1: num[0] < num[num.length - 1] 
        // if (num[lb] < num[ub]) return num[lb]; 
 
        // case2: num[0] > num[num.length - 1] or num[0] < num[num.length - 1] 
        while (lb + 1 < ub) { 
            int mid = lb + (ub - lb) / 2; 
            if (num[mid] < num[ub]) { 
                ub = mid; 
            } else if (num[mid] > num[ub]){ 
                lb = mid; 
            } else { 
                ub--; 
            } 
        } 
 
        return Math.min(num[lb], num[ub]); 
    } 
}

源码分析

注意num[mid] > num[ub]时应递减 ub 或者递增 lb.

复杂度分析

最坏情况下 O(n), 平均情况下 O(logn).

原创文章，作者：ItWorker，如若转载，请注明出处：https://blog.ytso.com/20652.html