[LeetCode] Self Crossing-优快云博客

本文介绍了一个一过性算法，该算法仅使用常数额外空间来确定由正数数组定义的路径是否交叉自身。通过三个具体例子说明了算法的应用场景，并详细解析了实现代码。

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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)

解题思路

相交可分为如下三种情况：

第四条线与第一条线相交
第五条线与第一条线相交或重叠
第六条线与第一条线相交

点Xi为给定的最后一个点。

实现代码

// Runtime: 1 ms
public class Solution {
    public boolean isSelfCrossing(int[] x) {
        int len = x.length;
        if(len <= 3) return false;

        for(int i = 3; i < len; i++) {
            if(x[i] >= x[i-2] && x[i-1] <= x[i-3]) {
                return true;
            }
            if(i >= 4) {
                if(x[i-1] == x[i-3] && x[i] + x[i-4] >= x[i-2]) {
                    return true;
                }
            }
            if(i >= 5) {
                if(x[i-2] - x[i-4] >= 0 && x[i] >= x[i-2] - x[i-4] && x[i-1] >= x[i-3] - x[i-5] && x[i-1] <= x[i-3]) {
                    return true; 
                }
            }
        }
        return false;
    }
}