题解 | #在旋转过的有序数组中寻找目标值#

在旋转过的有序数组中寻找目标值

https://www.nowcoder.com/practice/87c0e7abcbda41e7963660fa7d020995

import java.util.*;


public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
     *
     * 
     * @param nums int整型一维数组 
     * @param target int整型 
     * @return int整型
     */
    public int search (int[] nums, int target) {
        // write code here
        int ret = -1;
        int left = 0;
        int right = nums.length - 1;
        int mid = (left + right) / 2;
        //分两种情况
        if (nums[left] < nums[right]){
        //1.数组为升序
            while(left <= right){
                mid = (left + right) / 2;
                if (nums[mid] < target){
                    left = mid + 1;
                }else if (nums[mid] > target){
                    right = mid - 1;
                }else if (nums[mid] == target){
                    ret = mid;
                    break;
                }
            }
        }else{
        //2.数组被旋转过
            while(left <= right){
                mid = (left + right) / 2;
                System.out.println(nums[mid]);
                if (nums[mid] < target){
                    if (target > nums[right]){
                        right = mid - 1;
                    }else if (target < nums[right]){
                        left = mid + 1;
                    }else if (target == nums[right]){
                        ret = right;
                        break;
                    }
                }else if (nums[mid] > target){
                    if (target > nums[left]){
                        right = mid - 1;
                    }else if (target < nums[left]){
                        if (nums[mid] < nums[left]){
                            right = mid - 1;
                        }else{
                            left = mid + 1;
                        }
                    }else if (target == nums[left]){
                        ret = left;
                        break;
                    }
                }else if (nums[mid] == target){
                    ret = mid;
                    break;
                }
            }
        }

        return ret;
    }
}

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