一：用法
1、有序、无反复的数组
2、当有反复时，能够先用二分法查找到，而后应用左右滑动来找到边界
3、二分法例子：
- 给定一个n个元素有序的（升序）整型数组nums和一个目标值target，写一个函数搜寻nums中的target，如果目标值存在返回下标，否则返回-1。
public class Search { public static void main(String[] args) { int[] nums = {-1, 0, 3, 5, 9, 12};// int target = 9; int target = 2; //办法一：暴力循环 int i = method1(nums, target); System.out.println(i); //办法二：target【】双闭区间 int get2 = method2(nums, target); System.out.println(get2); //办法三：【）左闭右开区间 int get3 = method3(nums, target); System.out.println(get3); } private static int method3(int[] nums, int target) { int left = 0; int right = nums.length;//[) while (left < right) { int mid = left + ((right - left) >> 1); if (nums[mid] > target) { right = mid; } else if (nums[mid] < target) { left =mid+1; } else if (nums[mid] == target) { return mid; } } return -1; } private static int method2(int[] nums, int target) { int left = 0; int right = nums.length - 1; //因为此处是-1，所以要用双闭区间 while (left <= right) { int mid = left + ((right - left) >> 1); if (nums[mid] > target) { right = mid - 1; } else if (nums[mid] < target) { left = mid + 1; } else if (nums[mid] == target) { return mid; } } //未找到指标 return -1; } private static int method1(int[] nums, int target) { int get = -1; for (int i = 0; i < nums.length; i++) { if (nums[i] == target) { get = i; break; } } return get; }}