标签:.com https 图片 cal spel cloc minimum class step
In the video game Fallout 4, the quest "Road to Freedom" requires players to reach a metal dial called the "Freedom Trail Ring", and use the dial to spell a specific keyword in order to open the door.
Given a string ring, which represents the code engraved on the outer ring and another string key, which represents the keyword needs to be spelled. You need to find the minimum number of steps in order to spell all the characters in the keyword.
Initially, the first character of the ring is aligned at 12:00 direction. You need to spell all the characters in the string key one by one by rotating the ring clockwise or anticlockwise to make each character of the string key aligned at 12:00 direction and then by pressing the center button.
At the stage of rotating the ring to spell the key character key[i]:
Example:
Input: ring = "godding", key = "gd" Output: 4 Explanation: For the first key character ‘g‘, since it is already in place, we just need 1 step to spell this character. For the second key character ‘d‘, we need to rotate the ring "godding" anticlockwise by two steps to make it become "ddinggo". Also, we need 1 more step for spelling. So the final output is 4.
Note:
Approach #1: DP. [C++]
class Solution { public: int findRotateSteps(string ring, string key) { int N = ring.length(); int M = key.length(); int ans = INT_MAX; // dp[i][j] denote means the minimum of steps to spell the key[0:i] which key[i] at the index of j in ring. vector<vector<int>> dp(M, vector<int>(N, INT_MAX)); vector<vector<int>> char_idx(26, vector<int>()); for (int i = 0; i < N; ++i) { int c = ring[i] - ‘a‘; char_idx[c].push_back(i); } for (int i = 0; i < M; ++i) { int cur_char = key[i] - ‘a‘; for (int cur_pos : char_idx[cur_char]) { if (i == 0) { dp[i][cur_pos] = min(cur_pos, N - cur_pos) + 1; } else { int pre_char = key[i-1] - ‘a‘; for (int pre_pos : char_idx[pre_char]) { int diff = min(abs(cur_pos-pre_pos), N - abs(cur_pos-pre_pos)); dp[i][cur_pos] = min(dp[i][cur_pos], dp[i-1][pre_pos]+diff+1); //cout << dp[i][cur_pos] << endl; } } if (i == M-1) ans = min(ans, dp[M-1][cur_pos]); } } return ans; } };
Analysis:
https://leetcode.com/problems/freedom-trail/discuss/194601/DP-solution-with-detailed-text-and-video-explanation
标签:.com https 图片 cal spel cloc minimum class step
原文地址:https://www.cnblogs.com/ruruozhenhao/p/10340167.html