Circular Array Loop

update Sep 10, 2017 13:28

LeetCode

You are given an array of positive and negative integers. If a number n at an index is positive, then move forward n steps. Conversely, if it's negative (-n), move backward n steps. Assume the first element of the array is forward next to the last element, and the last element is backward next to the first element. Determine if there is a loop in this array. A loop starts and ends at a particular index with more than 1 element along the loop. The loop must be "forward" or "backward'.

Example 1: Given the array [2, -1, 1, 2, 2], there is a loop, from index 0 -> 2 -> 3 -> 0.

Example 2: Given the array [-1, 2], there is no loop.

Note: The given array is guaranteed to contain no element "0". Can you do it in O(n) time complexity and O(1) space complexity?

Basic Idea:

思路 1：slow-fast 用两个指针，一个slow一个fast，分别对正数和负数进行检查，相遇就有环。但是要另外记得处理环中只有一个元素的情况；

Java Code:

    class Solution {
        public boolean circularArrayLoop(int[] nums) {
            int len = nums.length;
            int positive = -1, negative = -1;
            // 对每个数字，检查circle，分为正数负数两种情况考虑
            for (int i = 0; i < len; ++i) {
                if (nums[i] > 0) {
                    if (checkCircle(nums, i)) return true;
                } else if (nums[i] < 0) {
                    if (checkCircle(nums, i)) return true;
                }
            }
            return false;
        }
        private boolean checkCircle(int[] nums, int start) {
            int len = nums.length;
            boolean positive = false;
            if (nums[start] > 0) positive = true;
            else positive = false;

            int slow = start, fast = start;
            while ((nums[slow] > 0) == positive && 
                   (nums[fast] > 0) == positive &&
                   (nums[getIndex(fast + nums[fast], len)] > 0) == positive) {
                slow = getIndex(slow + nums[slow], len);
                if (getIndex(slow + nums[slow], len) == slow) break; // 检查环中只有一个元素的情况
                fast = getIndex(fast + nums[fast], len); // 前进两步
                fast = getIndex(fast + nums[fast], len);
                if (nums[slow] == nums[fast]) return true;
            }
            return false;
        }
        // 需要注意负数取余需要另外操作；
        private int getIndex(int input, int len) {
            if (input >= 0) return input % len;
            return len - (len - input) % len;
        }
    }

思路 2：DFS 对每个数字进行dfs，如果有回路（visited），则说明有环，同样需要特别考虑环中只有一个元素的情况；

    class Solution(object):
        def circularArrayLoop(self, nums):
            """
            :type nums: List[int]
            :rtype: bool
            """
            def dfs(start, positive, visited):
                if (nums[start] > 0) != positive:
                    return False
                if start in visited: 
                    # check if there is only one element in circle
                    if (start + nums[start]) % len(nums) == start:
                        return False
                    return True

                visited.add(start)
                return dfs((nums[start] + start) % len(nums), positive, visited)


            for i in range(len(nums)):
                if nums[i] > 0:
                    if dfs(i, True, set()):
                        return True
                else:
                    if dfs(i, False, set()):
                        return True
            return False

update Dec 1, 2017 17:58

更新

因为Java对负数取余的时候和python不同，所以用Java写 slow-fast 的解法比较麻烦，用python 好了很多。

python slow-fast solution:

    class Solution(object):
        def circularArrayLoop(self, nums):
            """
            :type nums: List[int]
            :rtype: bool
            """
            def checkCircle(start):
                forward = True if nums[start] > 0 else False
                slow, fast = start, start
                while forward == (nums[slow] > 0) and \
                      forward == (nums[fast] > 0) and \
                      forward == (nums[(nums[fast] + fast) % len(nums)] > 0):
                    slow = (slow + nums[slow]) % len(nums)
                    fast = (fast + nums[fast]) % len(nums)
                    fast = (fast + nums[fast]) % len(nums)
                    if (slow + nums[slow]) % len(nums) == slow: return False
                    if slow == fast: return True
                return False

            # Main part
            for i in range(len(nums)):
                if nums[i] > 0 and checkCircle(i): return True
                elif nums[i] < 0 and checkCircle(i): return True
            return False

Java dfs solution:

    // dfs solution，正负情况分别讨论，遇到之前visited的就说明有环
    class Solution {
        public boolean circularArrayLoop(int[] nums) {
            for (int i = 0; i < nums.length; ++i) {
                if (nums[i] > 0 && dfs(nums, i, true, new HashSet<Integer>())) return true;
                if (nums[i] < 0 && dfs(nums, i, false, new HashSet<Integer>())) return true;
            }
            return false;
        }

        private boolean dfs(int[] nums, int curr, boolean isForward, Set<Integer> visited) {
            if (visited.contains(curr)) return true; // 有环
            if ((nums[curr] > 0) != isForward) return false; // 符号不对
            if (getIndex(curr + nums[curr], nums) == curr) return false; // 环中只有一个元素
            int next = getIndex(curr + nums[curr], nums);
            visited.add(curr);
            return dfs(nums, next, isForward, visited);
        }

        private int getIndex(int input, int[] nums) {
            int len = nums.length;
            if (input >= 0) return input % len;
            else return len - (len - input) % len;
        }
    }