问题分析

在旋转排序数组上应用二分查找，若数组中的元素都不相同，则可以按下图分析：

  • 若a[mid] > a[high], 则代表mid位于旋转数组的左半段；

  • 若a[mid] < a[low],则代表mid位于旋转数组的右半段

  • 否则，代表区间[low, high]是个递增数组，不存在旋转

若数组中可能存在重复元素，则需要按下图分析：

  • 若a[mid] < a[low], 则代表处于状态①
  • 若a[mid] > a[high], 则代表处于状态②
  • 若a[mid] = a[low] = a[high],则代表处于状态③或④
  • 否则，代表区间[low, high]不存在反转

1. 寻找旋转排序数组中的最小值

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public int findMin(int[] nums) {
    int low = 0, high = nums.length-1;

    while(low < high){
        int mid = low + (high - low) / 2;
        if(nums[mid] > nums[high]){
            low = mid+1;
        }
        else if(nums[mid] < nums[low]){
            high = mid;
        }
        else{
            high = mid;
        }
    }

    return nums[low];
}

2. 寻找旋转排序数组中的最小值II

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public int findMin(int[] nums) {

    int low = 0, high = nums.length-1;

    while(low < high){
        int mid = low + (high - low) / 2;
        if(nums[mid] < nums[low]){
            high = mid;
        }
        else if(nums[mid] > nums[high]){
            low = mid + 1;
        }
        else if(nums[mid] == nums[low] && nums[mid] == nums[high]){
            high--;
        }
        else{
            high = mid;
        }
    }

    return nums[low];


}

3. 搜索旋转排序数组

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public int search(int[] nums, int target) {

    int low = 0, high = nums.length-1;

    while(low <= high){
        int mid = low + (high - low) / 2;
        if(nums[mid] == target){
            return mid;
        }

        if(nums[mid] > nums[high]){
            if(target >= nums[low] && target < nums[mid]){
                high = mid - 1;
            }
            else{
                low = mid + 1;
            }
        }
        else if(nums[mid] < nums[low]){
            if(target > nums[mid] && target <= nums[high]){
                low = mid + 1;
            }
            else {
                high = mid - 1;
            }
        }
        else {
            if(target > nums[mid]){
                low = mid + 1;
            }
            else{
                high = mid - 1;
            }
        }
    }

    return -1;
}

4. 搜索旋转排序数组II

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public boolean search(int[] nums, int target) {

    int low = 0, high = nums.length-1;

    while(low <= high){
        int mid = low + (high - low) / 2;
        if(nums[mid] == target){
            return true;
        }

        if(nums[mid] > nums[high]){
            if(target >= nums[low] && target < nums[mid]){
                high = mid - 1;
            }
            else{
                low = mid + 1;
            }
        }
        else if(nums[mid] < nums[low]){
            if(target > nums[mid] && target <= nums[high]){
                low = mid + 1;
            }
            else{
                high = mid - 1;
            }
        }
        else if(nums[mid]  == nums[low] && nums[mid] == nums[high]){
            low++;
            high--;
        }
        else{
            if(target > nums[mid]){
                low = mid + 1;
            }
            else{
                high = mid - 1;
            }
        }

    }

    return false;
}