旋转数组中的最小数字，剑指offer，P70 二分查找来实现O（logn)的查找

Posted 2020-06-17

public class MinNumberInRotatedArray {

    public int getMinNumInRotatedArray(int[] array) {
        if(array == null) {
            return -1;
        }
        int leftIndex = 0;
        int rightIndex = array.length - 1;
        if(array[leftIndex] == array[rightIndex]) {
            return array[0];
        }
        int midIndex = leftIndex;
        while(array[leftIndex] >= array[rightIndex]) {
        //    while(true) { 换成这个其实一样
            if(rightIndex - leftIndex <= 1) {//循环退出的条件
                midIndex = rightIndex;
                break;
            }
            midIndex = (leftIndex+rightIndex)/2;
            if(array[leftIndex] == array[rightIndex] && array[leftIndex] == array[midIndex]) {//如果两边的值和中间的值相等，
                //则无法判断中间值在左边还是在右边，因此只能进行顺序查找
                minNumBySort(array, leftIndex, rightIndex);
            }
            if(array[leftIndex] <= array[midIndex]) {
                leftIndex = midIndex;
            }else if(array[rightIndex] >= array[midIndex]){
                rightIndex = midIndex;
            }
        }
        return array[midIndex];
    }
    
    public int minNumBySort(int[] arr, int leftIndex, int rightIndex) {
        int result = arr[leftIndex];
        for(int i = leftIndex; i < rightIndex; i++) {
            if(arr[i] < result) {
                result = arr[i];
            }
        }
        return result;
    }
    
    public static void main(String[] args) {
        MinNumberInRotatedArray testArray = new MinNumberInRotatedArray();
        int[] array = {4,5,6,1,2,3};
        System.out.println(testArray.getMinNumInRotatedArray(array));
    }
}