1496. Path Crossing

2020-06-29 23:06:42 阅读：361 来源： 互联网

标签：set ver get pos else 1496 path Crossing Path

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.

Return True if the path crosses itself at any point, that is, if at any time you are on a location you've previously visited. Return False otherwise.

Example 1:

Input: path = "NES"
Output: false 
Explanation: Notice that the path doesn't cross any point more than once.

Example 2:

Input: path = "NESWW"
Output: true
Explanation: Notice that the path visits the origin twice.

Constraints:

1 <= path.length <= 10^4
path will only consist of characters in {'N', 'S', 'E', 'W}

class Solution {
    public boolean isPathCrossing(String path) {
        List<Integer> pos = Arrays.asList(0,0);
        Set<List<Integer>> set = new HashSet();
        set.add(pos);
        // boolean cross = false;
        // System.out.println(pos.get(0) + " " + pos.get(1));
        for(char c: path.toCharArray()){
            int hor = pos.get(1);
            int ver = pos.get(0);
            if(c == 'N'){
                ver++;
            }
            else if(c == 'S'){
                ver--;
            }
            else if(c == 'E'){
                hor++;
            }
            else hor--;
            // pos = ;
            //System.out.println(pos[0] + " " + pos[1]);
            pos = Arrays.asList(ver, hor);
        // System.out.println(pos.get(0) + " " + pos.get(1));
            if(set.contains(pos)) return true;
            else set.add(pos);
        }
        return false;
    }
}

simulation，起始是0，0，把途径的点表示成arraylist加入set中看有没有重复

注意不能用array，因为就算两个数组元素相等，只要不是同1 reference也会判定不相等，而List的比较是也比较element。

class Solution {
    public boolean isPathCrossing(String path) {
        int x = 0, y = 0;
        Set<String> set = new HashSet();
        set.add(x + "$" + y);
        for (char c : path.toCharArray()) {
            if (c == 'N') y++;
            else if (c == 'S') y--;
            else if (c == 'E') x++;
            else x--;
            String key = x + "$" + y;
            if (set.contains(key))
                return true;
            set.add(key);
        }
        return false;
    }
}

像这种表示就更简单了，忘球了可惜，以前在哪里见过来着

标签：set,ver,get,pos,else,1496,path,Crossing,Path
来源： https://www.cnblogs.com/wentiliangkaihua/p/13210926.html

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1496. Path Crossing