33. Search in Rotated Sorted Array

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

(i.e.,0 1 2 4 5 6 7might become4 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.

解题方法

由連續區間去判斷： 畫圖！畫圖

public int search(int[] nums, int target) {
        //畫圖
        if(nums == null || nums.length == 0){
            return -1;
        }

        int start = 0;
        int end = nums.length - 1;
        while(start + 1 < end){
            int mid = start + (end -start) / 2;
            if(nums[mid] == target){
                return mid;
            }

            //左低右高
            //注意 <= && >=
            if(nums[start] < nums[mid]){
                if(nums[start] <= target && target < nums[mid]){
                    end = mid;
                }else{
                    start = mid;
                }
            }else{ //左高右低
                if(nums[mid] < target && target <= nums[end]){
                    start = mid; 
                }else{
                    end = mid;
                }
            }
        }

        if(nums[start] == target){
            return start;
        }else if(nums[end] == target){
            return end;
        }
        return -1;
    }

Follow up!

The array may contain duplicates.

解题方法

这个问题在面试中不会让实现完整程序

只需要举出能够最坏情况的数据是 [1,1,1,1... 1] 里有一个0即可。

在这种情况下是无法使用二分法的，复杂度是O(n)

因此写个for循环最坏也是O(n)，那就写个for循环就好了

如果你觉得，不是每个情况都是最坏情况，你想用二分法解决不是最坏情况的情况，那你就写一个二分吧。

反正面试考的不是你在这个题上会不会用二分法。这个题的考点是你想不想得到最坏情况。

多增加兩個case

else if( nums[start] == nums[mid]){
   start++;
}else{
    end--;
}

    public boolean search(int[] nums, int target) {
        //MergeSort
        if(nums == null || nums.length == 0){
            return false;
        }

        int start = 0;
        int end = nums.length - 1;
        while(start + 1 < end){
            int mid = start + (end - start) / 2;
            if(target == nums[mid]){
                return true;
            }

            if(nums[start] < nums[mid]){
                if(target >= nums[start] && target < nums[mid]){
                    end = mid;
                }else{
                    start = mid;
                }
            }else if(nums[mid] < nums[end]){
                if(nums[mid] < target && nums[end] >= target){
                    start = mid;
                }else{
                    end = mid;//duplicates
                }
            }else if( nums[start] == nums[mid]){
                start++;
            }else{
                end--;
            }
        }

        if(target == nums[start]){
            return true;
        }
        if(target == nums[end]){
            return true;
        }

        return false;
    }