题目地址：

https://leetcode.com/problems/minimum-sideway-jumps/description/

给定一条宽为333的路，可以视为三条平行线，给定一个长n+1n+1n+1的数组AAA，A[i]=0A[i]=0A[i]=0表示位置iii无障碍物，A[i]=k>0A[i]=k>0A[i]=k>0表示位置iii的第kkk条线有障碍物。一只青蛙在位置000的第二条平行线出发，可以以代价000向右跳一步，也可以以代价111变线，只是不能跳到障碍物上。题目保证A[0]=A[n]=0A[0]=A[n]=0A[0]=A[n]=0，问要跳到位置nnn的最小代价。

可以将所有位置看成是图上的点，代价视为边权，则代价只有0,10,10,1两种，从而是一个010101图上的最短路问题，可以用双端队列BFS来做，参考https://blog.csdn.net/qq_46105170/article/details/114667171。代码如下：

class Solution {public:using PII = pair<int, int>;int minSideJumps(vector<int>& obstacles) {int n = obstacles.size() - 1;deque<PII> dq;dq.push_back({0, 2});bool vis[n + 1][3];memset(vis, 0, sizeof vis);int dist[n + 1][3];memset(dist, 0x3f, sizeof dist);dist[0][1] = 0;while (dq.size()) {auto t = dq.front();dq.pop_front();int x = t.first, y = t.second;if (x == n) return dist[x][y - 1];if (vis[x][y - 1]) continue;vis[x][y - 1] = true;if (obstacles[x + 1] != y && !vis[x + 1][y - 1] &&dist[x + 1][y - 1] > dist[x][y - 1]) {dist[x + 1][y - 1] = dist[x][y - 1];dq.push_back({x + 1, y});}for (int k = 1; k <= 3; k++)if (k != y && k != obstacles[x] && !vis[x][k - 1] &&dist[x][k - 1] > dist[x][y - 1] + 1) {dist[x][k - 1] = dist[x][y - 1] + 1;dq.push_back({x, k});}}return -1;}
};

时空复杂度O(n)O(n)O(n)。

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【Leetcode】1824. Minimum Sideway Jumps

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