leetcode 81. 搜索旋转排序数组 II

假设按照升序排序的数组在预先未知的某个点上进行了旋转。

( 例如，数组 [0,0,1,2,2,5,6] 可能变为 [2,5,6,0,0,1,2] )。

编写一个函数来判断给定的目标值是否存在于数组中。若存在返回 true，否则返回 false。

示例 1:

输入: nums = [2,5,6,0,0,1,2], target = 0
输出: true
示例 2:

输入: nums = [2,5,6,0,0,1,2], target = 3
输出: false

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/search-in-rotated-sorted-array-ii
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

 

public class _81 {
    public boolean search(int[] nums, int target) {
        if (nums.length == 2) return nums[0] == target || nums[1] == target;
        if (nums.length == 1) return nums[0] == target;
        if (nums.length == 0) return false;
        int low = 0, high = nums.length-1, mid = (low+high)/2;
        int nsl = 0, nsr = high;

        if (nums[low] == nums[mid] && nums[mid] == nums[high]) {
            for (int i = 0; i < nums.length; i++)
                if (nums[i] == target)
                    return true;
            return false;
        }

        // 找出有序的那部分
        int[] t = null;
        if (nums[mid] <= nums[high]){ // 左边旋转过，右边有序
            t = Arrays.copyOfRange(nums, mid, high+1);
            nsl = low;
            nsr = mid+1;
        } else {
            t = Arrays.copyOfRange(nums, low, mid+1);
            nsl = mid;
            nsr = high+1;
        }
        int res = Arrays.binarySearch(t, target);
        if (res >= 0) return true;

        // 判断无序的左边是否存在target
        return search(Arrays.copyOfRange(nums, nsl, nsr), target);
    }

    public static void main(String[] args) {
        int[] nums = {1,1,1,1,1,1,1,2,1,1};
        boolean res = new _81().search(nums, 2);
        System.out.println("res = " + res);
    }
}

 稍微改进下。

public class _81 {
    public boolean binarySearch(int[] nums, int low, int high, int target){ // 包括high位置
        int mid;

        while(low <= high){
            mid = (low+high)/2;
            if (nums[mid] < target){
                low = mid+1;
            } else if (nums[mid] > target){
                high = mid - 1;
            } else {
                return true;
            }
        }
        return false;
    }
    public boolean result(int[] nums, int target, int low, int high){
        if (low+1==high) return nums[low] == target || nums[high] == target;
        if (low == high) return nums[low] == target;
        int mid = (low+high)/2;

        if (nums[low] == nums[mid] && nums[mid] == nums[high]) {
            for (int i = low; i <= high; i++)
                if (nums[i] == target)
                    return true;
            return false;
        }

        if (nums[mid] <= nums[high]){ // 右边有序，且taregt存在于右边
            if (target <= nums[high] && target >= nums[mid])
                return binarySearch(nums, mid, high, target);
            // 查找无序的左边
            return result(nums, target, low, mid);
        }
        if (nums[low] <= nums[mid]){ // 左边有序，且target存在于左边
            if (target <= nums[mid] && target >= nums[low])
                return binarySearch(nums, low, mid, target);
            // 查找无序的右边
            return result(nums, target, mid, high);
        }
        return false;
    }
    public boolean search(int[] nums, int target) {
        if (nums.length == 2) return nums[0] == target || nums[1] == target;
        if (nums.length == 1) return nums[0] == target;
        if (nums.length == 0) return false;
        return result(nums, target, 0, nums.length-1);
    }

    public static void main(String[] args) {
        int[] nums = {2,5,6,0,0,1,2};
        boolean res = new _81().search(nums, 8);
        System.out.println("res = " + res);
    }
}