Search in Rotated Sorted Array (M)

題目

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

題意

將一遞增數列的隨機后半部分與前半部分換位，得到新陣列，在新陣列中查找目標值，

思路

比較簡單的做法是先用一個二分找到兩個遞增序列的邊界，再用一個二分在對應的遞增序列里查找目標值，

也可以只用一個二分完成操作：求出mid，先考慮mid落在右遞增區間的情況，如果target也落在右遞增區間且比nums[mid]大，說明只需要繼續向右查找，令 left = mid + 1 即可，不然只要向左查找，令 right = mid - 1；同理，對于mid落在左遞增區間的情況進行處理，

代碼實作

Java

不求邊界

class Solution {
    public int search(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] == target) {
                return mid;
            }
            // 通過與第一個值相比較來判斷mid落在左區間還是右區間
            if (nums[mid] < nums[left]) {
                if (target <= nums[right] && target > nums[mid]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            } else {
                if (target >= nums[left] && target < nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            }
        }
        
        return -1;
    }
}

先求邊界

class Solution {
    public int search(int[] nums, int target) {
        if (nums.length == 0) {
            return -1;
        }
        
        int left = 0;
        int right = nums.length - 1;
        int split = nums.length - 1;		// 默認不存在邊界
        
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] == target) {
                return mid;
            }
            if (nums[mid] >= nums[left]) {
                if (mid + 1 < nums.length && nums[mid + 1] < nums[mid]) {
                    split = mid;
                    break;
                } else {
                    left = mid + 1;
                }
            } else {
                if (mid - 1 >= 0 && nums[mid - 1] > nums[mid]) {
                    split = mid - 1;
                    break;
                } else {
                    right = mid - 1;
                }
            }
        }
		
        // 選擇一個遞增區間進行二分查找
        if (target >= nums[0]) {
            return binarySearch(nums, 0, split, target);
        } else {
            return binarySearch(nums, split + 1, nums.length - 1, target);
        }
    }

    private int binarySearch(int[] nums, int left, int right, int target) {
        while (left <= right) {
            int mid = (left + right) / 2;
            if (target > nums[mid]) {
                left = mid + 1;
            } else if (target < nums[mid]) {
                right = mid - 1;
            } else {
                return mid;
            }
        }
        return -1;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function (nums, target) {
  let left = 0, right = nums.length - 1
  while (left <= right) {
    let mid = Math.trunc((right - left) / 2) + left

    if (nums[mid] === target) {
      return mid
    }

    if (target < nums[0] && nums[mid] >= nums[0]) {
      left = mid + 1
    } else if (target >= nums[0] && nums[mid] < nums[0]) {
      right = mid - 1
    } else if (nums[mid] < target) {
      left = mid + 1
    } else {
      right = mid - 1
    }
  }
  
  return -1
}

標籤：其他

上一篇：vlan間互通的奇怪問題。

下一篇：0275. H-Index II (M)