#32 动态规划: 最长有效括号对长度问题 #20
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azl397985856
approved these changes
Jun 4, 2019
| 2. s的前**1**个子字符串的最长有效括号对长度为0, | ||
| s的前**2**个子字符串的最长有效括号对长度也为0, | ||
| 这个时候我们可以得出结论: 最长有效的括号对不可能以'('结尾; | ||
| 3. 当i等于3时, 我们可以看出dp(2)为0, dp(3)=2, 因为第2个字符和第3个字符是配对的, |
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| ## 扩展 | ||
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| 1. 如果判断的不仅仅只有(和), 还有[, ], {和}, 改怎么办? |
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字符左右建议加引号， 比如"[", "]"
改怎么办 改为 “该怎么办”
| ## 扩展 | ||
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| 1. 如果判断的不仅仅只有(和), 还有[, ], {和}, 改怎么办? | ||
| 2. 如果输出的不是长度, 而是最长有效括号对的字符串, 改怎么办? |
| 这个时候我们可以得出结论: 最长有效的括号对不可能以'('结尾; | ||
| 3. 当i等于3时, 我们可以看出dp(2)为0, dp(3)=2, 因为第2个字符和第3个字符是配对的, | ||
| 当i等于4时, dp(i-1)为2, dp(4)为4, 我们配对的是第1个字符和第4个字符, | ||
| 因此, 我们可以得出结论: 如果第**i**个字符和第**i-1-dp(i-1)**个字符是配对的, 则dp(i) = dp(i-1) + 2; |
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3 4 5 的描述有点啰嗦。 建议说出思路即可，最好配个图
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