43 lines
1.2 KiB
Markdown
43 lines
1.2 KiB
Markdown
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//dp[i][j]，key的0~i位字符拼写后，ring的第j位对齐12：00方向，需要的最小步数
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//前提：key[i] = ring[j]，若不满足，dp[i][j] = INT_MAX
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这道题目我服！ 没做出来
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https://blog.csdn.net/qq_41855420/article/details/89058979
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```
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class Solution {
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public:
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int findRotateSteps(string ring, string key) {
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//int dp[101][101] = {0};
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int n = ring.size();
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vector<vector<int>> dp(key.size() + 1, vector<int>(ring.size(), 0));
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for (int i = key.size() - 1; i >= 0; i--) {
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for (int j = 0; j < ring.size(); j++) {
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dp[i][j] = INT_MAX;
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for (int k = 0; k < ring.size(); k++) {
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if (ring[k] == key[i]) {
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int diff = abs(j - k);
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int step = min(diff, n - diff);
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dp[i][j] = min(dp[i][j], step + dp[i + 1][k]);
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}
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}
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}
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}
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for (int i = 0; i < dp.size(); i++) {
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for (int j = 0; j < dp[0].size(); j++) {
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cout << dp[i][j] << " ";
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}
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cout << endl;
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}
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return dp[0][0] + key.size();
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}
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};
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```
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2 3 4 5 5 4 3
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2 1 0 0 1 2 3
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