Leetcode 657 机器人能否回到原点

  JAVA 遍历:

public final boolean judgeCircle(String moves) {
        int x = 0, y = 0;
        for (int i = 0; i < moves.length(); i++) {
            char c = moves.charAt(i);
            switch (c) {
                case 'L':
                    x--;
                    break;
                case 'R':
                    x++;
                    break;
                case 'U':
                    y++;
                    break;
                case 'D':
                    y--;
                    break;
            }
        }
        return x == 0 && y == 0;
    }

  JAVA 栈辅助:

public final boolean judgeCircle(String moves) {
        Stack<Character> uStack = new Stack<Character>();
        Stack<Character> dStack = new Stack<Character>();
        Stack<Character> lStack = new Stack<Character>();
        Stack<Character> rStack = new Stack<Character>();
        for (int i = 0; i < moves.length(); i++) {
            Stack<Character> t = null, s = null;
            switch (moves.charAt(i)) {
                case 'L':
                    t = rStack;
                    s = lStack;
                    break;
                case 'R':
                    t = lStack;
                    s = rStack;
                    break;
                case 'U':
                    t = dStack;
                    s = uStack;
                    break;
                case 'D':
                    t = uStack;
                    s = dStack;
                    break;
            }
            popOrPush(moves.charAt(i), t, s);
        }
        return uStack.empty() && dStack.empty() && lStack.empty() && rStack.empty();
    }

    private final void popOrPush(char c, Stack<Character> t, Stack<Character> s) {
        if (t.empty()) s.push(c);
        else t.pop();
    }

  JS 栈辅助:

/**
 * @param {string} moves
 * @return {boolean}
 */
var judgeCircle = function (moves) {
    var u = [], d = [], l = [], r = [];
    for (let i = 0; i < moves.length; i++) {
        let c = moves.charAt(i);
        popOrPush(c, u, d, l, r);
    }
    return u.length == 0 && d.length == 0 && l.length == 0 && r.length == 0;
}

var popOrPush = (c, u, d, l, r) => {
    let stack0, stack1;
    switch (c) {
        case 'U':
            stack0 = d;
            stack1 = u;
            break;
        case 'D':
            stack0 = u;
            stack1 = d;
            break;
        case 'L':
            stack0 = r;
            stack1 = l;
            break;
        case 'R':
            stack0 = l;
            stack1 = r;
    }
    if (stack0.length == 0) stack1.push(c);
    else stack0.pop();
}

 

posted @ 2020-12-01 23:02  牛有肉  阅读(115)  评论(0编辑  收藏  举报