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‎README.adoc‎

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//|Easy
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//|
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//|{counter:codes}
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//|{leetcode_base_url}/path-crossing/[1496. Path Crossing^]
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//|{source_base_url}/_1496_PathCrossing.java[Java]
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//|{doc_base_url}/1496-path-crossing.adoc[题解]
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//|Easy
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//|
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|{counter:codes}
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|{leetcode_base_url}/path-crossing/[1496. Path Crossing^]
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|{source_base_url}/_1496_PathCrossing.java[Java]
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|{doc_base_url}/1496-path-crossing.adoc[题解]
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|Easy
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|
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//|{counter:codes}
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//|{leetcode_base_url}/check-if-array-pairs-are-divisible-by-k/[1497. Check If Array Pairs Are Divisible by k^]

‎docs/1496-path-crossing.adoc‎

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[#1496-path-crossing]
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= 1496. Path Crossing
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= 1496. 判断路径是否相交
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{leetcode}/problems/path-crossing/[LeetCode - 1496. Path Crossing ^]
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https://leetcode.cn/problems/path-crossing/[LeetCode - 1496. 判断路径是否相交 ^]
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Given a string `path`, where `path[i] = 'N'`, `'S'`, `'E'` or `'W'`, each representing moving one unit north, south, east, or west, respectively. You start at the origin `(0, 0)` on a 2D plane and walk on the path specified by `path`.
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给你一个字符串 `path`，其中 `path[i]` 的值可以是 `N`、`S`、`E` 或者 `W`，分别表示向北、向南、向东、向西移动一个单位。
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Return `true` _if the path crosses itself at any point, that is, if at any time you are on a location you have previously visited_. Return `false` otherwise.
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你从二维平面上的原点 `(0, 0)` 处开始出发，按 `path` 所指示的路径行走。
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*Example 1:*
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<img alt="" src="https://assets.leetcode.com/uploads/2020/06/10/screen-shot-2020-06-10-at-123929-pm.png" style="width: 400px; height: 358px;" />
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[subs="verbatim,quotes"]
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----
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*Input:* path = "NES"
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*Output:* false
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*Explanation:* Notice that the path doesn't cross any point more than once.
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如果路径在任何位置上与自身相交，也就是走到之前已经走过的位置，请返回 `true` ；否则，返回 `false` 。
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----
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*示例 1：*
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*Example 2:*
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<img alt="" src="https://assets.leetcode.com/uploads/2020/06/10/screen-shot-2020-06-10-at-123843-pm.png" style="width: 400px; height: 339px;" />
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[subs="verbatim,quotes"]
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----
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*Input:* path = "NESWW"
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*Output:* true
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*Explanation:* Notice that the path visits the origin twice.
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----
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image::images/1496-01.png[{image_attr}]
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....
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输入：path = "NES"
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输出：false
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解释：该路径没有在任何位置相交。
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....
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*Constraints:*
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*示例 2：*
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image::images/1496-02.png[{image_attr}]
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* `1 <= path.length <= 10^4^`
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* `path[i]` is either `'N'`, `'S'`, `'E'`, or `'W'`.
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....
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输入：path = "NESWW"
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输出：true
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解释：该路径经过原点两次。
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....
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*提示：*
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* `1 \<= path.length \<= 10^4^`
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* `path[i]` 为 `N`、`S`、`E` 或 `W`
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== 思路分析
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想找个巧办法，结果失败！还是用记录轨迹的办法搞定了。
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因为题目要求 `1 \<= path.length \<= 10^4^`，那么 `x` 和 `y` 的值，最大不会超过 `10^4^`，直接讲 `(x, y)` 转换成 `x * 10000 + y` 一个数字来作为坐标即可。
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[[src-1496]]
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[tabs]
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== 参考资料
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. https://leetcode.cn/problems/path-crossing/solutions/310426/pan-duan-lu-jing-shi-fou-xiang-jiao-by-leetcode-so/[1496. 判断路径是否相交 - 官方题解^]

‎docs/images/1496-01.png‎

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‎docs/images/1496-02.png‎

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‎docs/index.adoc‎

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// include::1495-friendly-movies-streamed-last-month.adoc[leveloffset=+1]
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// include::1496-path-crossing.adoc[leveloffset=+1]
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include::1496-path-crossing.adoc[leveloffset=+1]
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// include::1497-check-if-array-pairs-are-divisible-by-k.adoc[leveloffset=+1]
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‎logbook/202503.adoc‎

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|{doc_base_url}/1387-sort-integers-by-the-power-value.adoc[题解]
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|✅ 先计算每个数字的权重，再进行比较：使用集合存储下标，通过下标找到对应的权重。计算权重时，可以递归，也可以直接循环。使用递归时，可以使用备忘录把已经计算的值存储起来，防止重复计算。
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|{counter:codes2503}
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|{leetcode_base_url}/path-crossing/[1496. 判断路径是否相交^]
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|{doc_base_url}/1496-path-crossing.adoc[题解]
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|✅ 想找个巧办法，结果失败！还是用记录轨迹的办法搞定了。
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|===
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截止目前，本轮练习一共完成 {codes2503} 道题。

‎src/main/java/com/diguage/algo/leetcode/_1496_PathCrossing.java‎

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package com.diguage.algo.leetcode;
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import java.util.HashSet;
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import java.util.Set;
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public class _1496_PathCrossing {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2025-05-27 20:32:05
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*/
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public boolean isPathCrossing(String path) {
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int x = 0, y = 0;
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Set<Integer> visited = new HashSet<>();
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visited.add(0);
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for (int i = 0; i < path.length(); i++) {
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char c = path.charAt(i);
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if (c == 'N') {
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y++;
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} else if (c == 'S') {
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y--;
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} else if (c == 'E') {
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x++;
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} else if (c == 'W') {
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x--;
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}
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int point = x * 10000 + y;
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if (visited.contains(point)) {
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return true;
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} else {
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visited.add(point);
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}
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}
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return false;
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}
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// end::answer[]
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}

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