LeetCode-Self Crossing-优快云博客

本文介绍了一种O(1)额外空间复杂度的一次遍历算法，用于判断给定路径是否会发生交叉。通过分析路径的移动方向和距离，算法能够准确地在不使用额外空间的情况下确定路径是否交叉。

You are given an array x of n positive numbers. You start at point (0,0) and moves x[0] metres to the north, then x[1] metres to the west, x[2] metres to the south,x[3] metres to the east and so on. In other words, after each move your direction changes counter-clockwise.

Write a one-pass algorithm with O(1) extra space to determine, if your path crosses itself, or not.

Example 1:

Given x = [2, 1, 1, 2],
┌───┐
│   │
└───┼──>
    │

Return true (self crossing)
Example 2:

Given x = [1, 2, 3, 4],
┌──────┐
│      │
│
│
└────────────>

Return false (not self crossing) 
Example 3:

Given x = [1, 1, 1, 1],
┌───┐
│   │
└───┼>

Return true (self crossing)

这题考验逻辑归类能力。看了别人的思路，其实只有三种情况能crossing

如果在一个轮回内，情况比较简单，只有一种可能crossing。

如果超过了一个轮回，有两种情况。

 1 public class Solution {
 2     public boolean isSelfCrossing(int[] x) {
 3         int len=x.length;
 4         //boolean b=false;
 5         if(len < 4){
 6             return false;
 7         }
 8         else{
 9             for(int i=3; i<len; i++){
10                 if(x[i-3]>=x[i-1] && x[i-2]<=x[i]){
11                    return true;
12                 }
13                 if(i>=4 && x[i-1]==x[i-3] && x[i]+x[i-4]>=x[i-2]){
14                     return true;
15                 }
16                 if(i>=5 && i >= 5 && x[i-2] >= x[i-4] && x[i-3] >= x[i-1] && x[i-1] >= x[i-3] - x[i-5] && x[i] >= x[i-2] - x[i-4]){
17                     return true;
18                 }
19             }
20         }
21         return false;
22     }
23 }