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;
}
}