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  • Find Minimum in Rotated Sorted Array II

    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).

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  • 原文地址:https://www.cnblogs.com/lyc94620/p/12778677.html
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