基于二分法
index1为首，index2为尾，indexMid指向中间
当Number[index1]大于等于Number[index2]的条件满足时
判断index2和index1的差距是否等于1
如果相等，说明index2即为那个突变点，最小值，将index2赋给indexMid，最终返回Number[indexMid]
如果不等，就取indexMid=index1和index2的中间值
如果index1，index2和indexMid相等的话，那么就需要在index1和index2之间用顺序查找
如果不等，若index1指向的数小于indexMid指向的数，将index1更新为indexMid
若indexMId指向的数小于index2指向的数，将index2更新为indexMid
顺序查找
相邻两个数比较，如果前一个数大于后一个数，那么后一个数就是最小值。如果不是，则往后找
package minNumberInRotatedArray8;
public class OriCode_MinInRotatedNumber {
public static void main(String[] args) {
// int[] numbers = { 3, 4, 5, 1, 2 };
// int[] numbers = { 1, 0, 1, 1, 1 };
int[] numbers={1,1,1,0,1};
int result = Min(numbers);
System.out.println(result);
}
static int Min(int[] numbers) {
int index1 = 0;
int index2 = numbers.length – 1;
int indexMid = index1;
while (numbers[index1] >= numbers[index2]) {
if (index2 – index1 == 1) {
indexMid = index2;
break;
}
indexMid = (index1 + index2) / 2;
if (numbers[index1] == numbers[index2]
&& numbers[index1] == numbers[indexMid]) {
return MinInOrder(numbers, index1, index2);
}
if (numbers[indexMid] >= numbers[index1]) {
index1 = indexMid;
} else if (numbers[indexMid] <= numbers[index2]) {
index2 = indexMid;
}
}
return numbers[indexMid];
};
static int MinInOrder(int[] numbers, int index1, int index2) {
int result = numbers[index1];
for (int i = index1 + 1; i < index2; i++) {
if (result > numbers[i]) {
result = numbers[i];
}
}
return result;
}
}
