Leetcode (Sword finger offer) - 11. Minimum number of rotation array

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The main idea of the topic ： A little

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AC Code

• Java

//  Solution (1)
class Solution {
    public int minArray(int[] numbers) {
        int min = Math.min(numbers[0], numbers[numbers.length - 1]);
        for (int i = 1; i < numbers.length - 1; i++) {
            if (numbers[i] >= numbers[i - 1] && numbers[i] > numbers[i + 1]) {
                min = numbers[i + 1];
                break;
            } else if (numbers[i] < numbers[i - 1]) {
                min = numbers[i];
                break;
            }
        }
        return min;
    }
}

//  Solution (2)
class Solution {
    public int minArray(int[] numbers) {
        int i = 0, j = numbers.length - 1;
        while (i < j) {
            int m = (i + j) / 2;
            if (numbers[m] > numbers[j]) i = m + 1;
            else if (numbers[m] < numbers[j]) j = m;
            else {
                int x = i;
                for(int k = i + 1; k < j; k++) {
                    if(numbers[k] < numbers[x]) x = k;
                }
                return numbers[x];
            }
        }
        return numbers[i];
    }
}

• C++

class Solution {
public:
    int minArray(vector<int>& numbers) {
        int i = 0, j = numbers.size() - 1;
        while (i < j) {
            int m = (i + j) / 2;
            if (numbers[m] > numbers[j]) i = m + 1;
            else if (numbers[m] < numbers[j]) j = m;
            else {
                int x = i;
                for(int k = i + 1; k < j; k++) {
                    if(numbers[k] < numbers[x]) x = k;
                }
                return numbers[x];
            }
        }
        return numbers[i];
    }
};