LeetCode 2833.距离原点最远的点：计数

【LetMeFly】2833.距离原点最远的点：计数

力扣题目链接：https://leetcode.cn/problems/furthest-point-from-origin/

给你一个长度为 n 的字符串 moves ，该字符串仅由字符 'L'、'R' 和 '_' 组成。字符串表示你在一条原点为 0 的数轴上的若干次移动。

你的初始位置就在原点（0），第 i 次移动过程中，你可以根据对应字符选择移动方向：

如果 moves[i] = 'L' 或 moves[i] = '_' ，可以选择向左移动一个单位距离
如果 moves[i] = 'R' 或 moves[i] = '_' ，可以选择向右移动一个单位距离

移动 n 次之后，请你找出可以到达的距离原点最远的点，并返回 从原点到这一点的距离 。

示例 1：

复制代码

输入：moves = "L_RL__R"
输出：3
解释：可以到达的距离原点 0 最远的点是 -3 ，移动的序列为 "LLRLLLR" 。

示例 2：

复制代码

输入：moves = "_R__LL_"
输出：5
解释：可以到达的距离原点 0 最远的点是 -5 ，移动的序列为 "LRLLLLL" 。

示例 3：

复制代码

输入：moves = "_______"
输出：7
解释：可以到达的距离原点 0 最远的点是 7 ，移动的序列为 "RRRRRRR" 。

提示：

1 <= moves.length == n <= 50
moves 仅由字符 'L'、'R' 和 '_' 组成

解题方法：计数

凡是L和R皆为固定，凡是其他皆自由。

因此两个变量统计：

L和R作用下距离起点的diff；
有多少自由移动的步数。

如果diff大于0（在起点之右），则自由步数全部向右移动；否则全部向左移动就好了。

此外，也可以不判断diff是否大于零，直接返回 a b s ( d i f f ) + f l e x abs(diff)+flex abs(diff)+flex就好了。

时间复杂度 O ( l e n ( m o v e s ) ) O(len(moves)) O(len(moves))
空间复杂度 O ( 1 ) O(1) O(1)

AC代码

C++

cpp 复制代码

/*
 * @LastEditTime: 2026-04-24 21:00:44
 */
class Solution {
public:
    int furthestDistanceFromOrigin(string moves) {
        int flex = 0, diff = 0;
        for (char c : moves) {
            switch (c)
            {
            case 'L':
                diff--;
                break;
            case 'R':
                diff++;
                break;
            default:
                flex++;
            }
        }
        return diff > 0 ? diff + flex : flex - diff;
    }
};

Python

python 复制代码

'''
LastEditTime: 2026-04-24 21:02:38
'''
class Solution:
    def furthestDistanceFromOrigin(self, moves: str) -> int:
        flex = diff = 0
        for c in moves:
            if c == 'L':
                diff -= 1
            elif c == 'R':
                diff += 1
            else:
                flex += 1
        return flex + diff if diff > 0 else flex - diff

Java

java 复制代码

/*
 * @LastEditTime: 2026-04-24 21:07:01
 */
// java14+
class Solution {
    public int furthestDistanceFromOrigin(String moves) {
        int flex = 0, diff = 0;
        for (int i = 0; i < moves.length(); i++) {
            switch (moves.charAt(i)) {
                case 'L' -> diff--;
                case 'R' -> diff++;
                default -> flex++;
            }
        }
        return diff > 0 ? flex + diff : flex - diff;
    }
}

Go

go 复制代码

/*
 * @LastEditTime: 2026-04-24 21:25:20
 */
package main

func furthestDistanceFromOrigin(moves string) int {
    flex, diff := 0, 0
    for _, c := range moves {
        switch byte(c) {
        case 'L':
            diff--
        case 'R':
            diff++
        default:
            flex++
        }
    }
    if diff > 0 {
        return diff + flex
    }
    return flex - diff
}

Rust - abs

rust 复制代码

/*
 * @LastEditTime: 2026-04-24 23:09:59
 */
impl Solution {
    pub fn furthest_distance_from_origin(moves: String) -> i32 {
        let mut diff: i32 = 0;
        let mut flex: i32 = 0;
        for c in moves.chars() {
            match c {
                'L' => diff -= 1,
                'R' => diff += 1,
                _ => flex += 1,
            }
        }
        diff.abs() + flex
    }
}

Rust - if-else

rust 复制代码

/*
 * @LastEditTime: 2026-04-24 23:07:48
 */
impl Solution {
    pub fn furthest_distance_from_origin(moves: String) -> i32 {
        let mut diff = 0;
        let mut flex = 0;
        for c in moves.chars() {
            match c {
                'L' => diff -= 1,
                'R' => diff += 1,
                _ => flex += 1,
            }
        }
        if diff > 0 {
            diff + flex
        } else {
            flex - diff
        }
    }
}

同步发文于CSDN和我的个人博客，原创不易，转载经作者同意后请附上原文链接哦~

千篇源码题解已开源