457. Circular Array Loop

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原文链接：https://my.oschina.net/liufq/blog/2249628

本文探讨了一个中等难度的算法问题：如何在O(n)时间复杂度和O(1)空间复杂度下，判断一个正负整数混合的数组是否存在循环。通过设定方向规则和遍历数组，文章详细解释了循环检测的实现过程。

为什么80%的码农都做不了架构师？>>>

Description

Difficulty : Medium

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

Points to be cared

A loop starts and ends at a particular index with more than 1 element along the loop. It means a satisfied loop contains more than 1 element
That the loop must be "forward" or "backward' means in a satisfied loop elements must be all postive or all negative

Mindpath:
Since the give array is guaranteed to contain no 0, we start from a index which is not zero then loop through the array.

Calculate next index and do next steps

check next index is zero or not, if it is, break
set next index element to zero. It means that we visited that element
check the direction, if changed, break
check whether next index is equal the initial index, if so, we find a loop

Code blow:

func circularArrayLoop(nums []int) bool {
    size := len(nums)
    if size <= 1 {
        return false
    }

    for i:=0;i<size;i++{
        if nums[i] == 0{
            continue
        }
        direction := nums[i] > 0
        curIndex := i
        curVal := nums[i]
        counter := 1
        for true{
            curIndex = curIndex + curVal
            if curIndex >= size {
                curIndex = curIndex - size
            } else if curIndex < 0 {
                curIndex = curIndex + size
            }
            curVal = nums[curIndex]
            //traveled
            if nums[curIndex] == 0 {
                break
            }
            nums[curIndex] = 0
            //loop detected
            if curIndex == i &&  counter > 1{
                return true
            }
            // direction changed
            if !((direction && curVal >0) || (!direction && curVal < 0)){
                break
            }
            counter++
        }
    }
    
    return false
}

转载于:https://my.oschina.net/liufq/blog/2249628