[LeetCode] 457. Circular Array Loop

最新推荐文章于 2019-11-13 20:03:25 发布

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最新推荐文章于 2019-11-13 20:03:25 发布

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原文链接：https://segmentfault.com/a/1190000017048008

本文探讨了如何在正负整数构成的数组中检测是否存在循环，重点在于使用O(n)时间复杂度和O(1)空间复杂度实现这一目标。通过慢快指针法和特殊数组遍历策略，文章提供了一种高效且简洁的解决方案。

Problem

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?

Solution

class Solution {
    public boolean circularArrayLoop(int[] nums) {
        if (nums == null || nums.length <= 2) return false;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 0) return false;
            int slow = i, fast = getIndex(nums, i);
            while (nums[fast] * nums[i] > 0 && nums[getIndex(nums, fast)] * nums[i] > 0) {
                if (slow == fast) {
                    if (slow == getIndex(nums, slow)) break;
                    return true;
                }
                slow = getIndex(nums, slow);
                fast = getIndex(nums, getIndex(nums, fast));
            }
            slow = i;
            int pre = nums[i];
            while (nums[slow] * pre > 0) {
                int next = getIndex(nums, slow);
                nums[slow] = 0;
                slow = next;
            }
        }
        return false;
    }
    private int getIndex(int[] nums, int i) {
        int len = nums.length;
        int next = i+nums[i];
        return next >= 0 ? next%len : next%len+len;
    }
}