剑指offer 8. 旋转数组的最小数字

// 题目：输入一个旋转数组，输出其中的最小值// 解法：使用二分法进行查找public class Main {public static void main(String[] args) {System.out.println(findMinNum(new int[]{1,0,1,1,1}));}public static int findMinNum(int[] input){int low = 0;int high = input.length-1;int mid = low;while(input[low]>=input[high]){if(high-low == 1){//如果high-low = 1，则返回high的值，因为high的值一定小于等于low的mid = high;break;}mid = (low+high)/2;if(input[low] == input[high] && input[mid] == input[low]){//如果出现low，mid，high都相等，则只能顺序查找return minInOrder(input, low, high);}if(input[mid]>=input[low]){//如果mid>=low则最小值在mid和high之间low = mid;}if(input[mid]<=input[high]){//如果mid<high则最小值在low和mid之间high = mid;}}return input[mid];}public static int minInOrder(int[] input, int low, int high){//顺序查找函数int result = input[low];for(int i = low+1;i<=high;i++){if(input[i]<result){result = input[i];}}return result;}}