四种二分查找法模板

public class BinarySearch {
    public static void main(String[] args) {
        int[] arr = {1,2,3,3,3,4,5,5,7};
        int upper1 = floor_lower(arr, 3);
        System.out.println(upper1);

    }
    //1.寻找第一个大于 target 的元素的下标，本案例中4 就是第一个大于 target 的元素 ，下标 = 5。
    public static int upper(int[] arr,int target){
        int left = 0;
        int right = arr.length-1;
        if(arr[right] <= target) return -1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (arr[mid] > target) {
                //[l,mid,r]
                right = mid;
            } else if (arr[mid] == target) {
                left = mid + 1;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
    //2.如果数组中存在元素等于 target，则返回最后一个等于target 的元素下标，如果不存在，则返回第一个大于 target 的元素下标。本案例中最后一个等于target的下标 = 4。
    public static int  floor_upper(int[] arr,int target){
        int left = 0;
        int right = arr.length-1;
        if(arr[right] < target) return -1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (arr[mid] > target) {
                //[l,mid,r]
                right = mid;
            } else if (arr[mid] == target) {
                left = mid + 1;
            } else {
                left = mid + 1;
            }
        }
        if (left - 1 >= 0 && arr[left - 1] == target){
            return  left - 1;
        }
        return left;
    }
    //3.寻找最后一个小于 target 的元素的下标，本案例中 2 则是最后一个小于 target 的，下标 = 1.
    public static int lower(int[] arr,int target){
        int left = 0;
        int right = arr.length - 1;
        if(arr[left] >= target) return -1;
        while (left < right){
            int mid = left+(right-left+1)/2;
            if (arr[mid] < target){
                left = mid;
            }else if(arr[mid] == target){
                right = mid - 1;
            }else {
                right = mid - 1;
            }
        }
        return right;
    }
    //4.如果数组中存在元素等于 target，则返回第一个等于target 的下标，如果不存在，则返回最后一个小于 target 的元素的下标。本案例中第一个等于target的下标 = 2。
    public static int floor_lower(int[] arr,int target){
        int left = 0;
        int right = arr.length - 1;
        if(arr[left] > target) return -1;
        while (left < right){
            int mid = left+(right-left+1)/2;
            if (arr[mid] < target){
                left = mid;
            }else if(arr[mid] == target){
                right = mid - 1;
            }else {
                right = mid - 1;
            }
        }
        if (right + 1 < arr.length && arr[right + 1] == target) return right + 1;
        return right;
    }
}