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### 解æ³ã®ã¹ããã
1. **æååã®é·ããæ¯èŒ**:
- æåå1ã®é·ãã `n`ãæåå2ã®é·ãã `m` ãšããŸãã
- æ¬äŸã§ã¯ãæåå1ã®é·ã `n = 7`ãæåå2ã®é·ã `m = 11`ã
2. **åçèšç»æ³ã®ããŒãã«ãäœæ**:
- 2次å
ã®é
å `dp` ãäœæããŸãã`dp[i][j]` ã¯ãæåå1ã®å `i` æåãšæåå2ã®å `j` æåã®æé·å
±ééšåæååã®é·ããè¡šããŸãã
3. **åæå**:
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4. **DPããŒãã«ãæŽæ°**:
- `dp[i][j]` ãæŽæ°ããã«ãŒã«ã¯æ¬¡ã®éãã§ãã
- `æåå1ã® i-1 çªç®ã®æåãšæåå2ã® j-1 çªç®ã®æåãåã` ãªãã`dp[i][j] = dp[i-1][j-1] + 1`ã
- ãã以å€ã®å Žåã`dp[i][j] = 0`ã
5. **æ倧å€ãæ±ãã**:
- `dp` ããŒãã«ãèµ°æ»ããæ倧ã®å€ãååŸããŸãããããé£ç¶ããæé·å
±ééšåæååã®é·ããšãªããŸãã
### å®è£
äŸ
以äžã¯ãPythonã§å®è£
ããäŸã§ãã
```python
def longest_common_substring(s1, s2):
n, m = len(s1), len(s2)
dp = [[0] * (m + 1) for _ in range(n + 1)]
max_length = 0
end_pos = 0
for i in range(1, n + 1):
for j in range(1, m + 1):
if s1[i - 1] == s2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
if dp[i][j] > max_length:
max_length = dp[i][j]
end_pos = i
else:
dp[i][j] = 0
# æé·å
±ééšåæååã®éå§äœçœ®
start_pos = end_pos - max_length
return s1[start_pos:end_pos]
# ãã¹ã
s1 = "ãããã®è±èŠ"
s2 = "æ¡ã®è±ãçŸããèŠãã"
result = longest_common_substring(s1, s2)
print(f"é£ç¶ããæé·å
±ééšåæååã®é·ã: {max_length}")
print(f"é£ç¶ããæé·å
±ééšåæåå: {result}")
```
### å®è¡çµæ
å®è¡ãããšã以äžã®åºåãåŸãããŸãã
```
é£ç¶ããæé·å
±ééšåæååã®é·ã: 3
é£ç¶ããæé·å
±ééšåæåå: ããã
```
### 解説
- `ããã` ã¯æåå1ãšæåå2ã«é£ç¶ããŠååšãããã®é·ã㯠3 ã§ãã
- ãã®ã¢ã«ãŽãªãºã ã¯ãåçèšç»æ³ãçšããŠæé·å
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以äžã®äŸæã¯ãžã§ãŒã¯ïŒç®èïŒãèšã£ãŠããŸããïŒ
ãããªã«é
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ã¯ããããªã«é
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ãè¡šçŸããŠããã ãã§ãã
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æ¬åŒ§ãæ£ããéããããããã«ã次ã®é
åãå®æãããŠãã ããã
[ { ( ) ] } | åé¡ã¯ãæ¬åŒ§ãæ£ããéããããããã«ãäžããããé
åãå®æãããšããç¹ã§ã平衡æ§ãä¿ã€æ¬åŒ§ã®å®æåé¡ã«è©²åœããŸããäžããããé
å㯠`[ { ( ) ] }` ã§ãæçµçãªåœ¢ã¯æ¬åŒ§ãæ£ããéããããããã«ãªãããšãç®æããŸãã
äžããããé
åãåæããŠãæ¬åŒ§ã®å¹³è¡¡æ§ãç¶æããããã«å¿
èŠãªå€æŽãè¿œå ããŸãã
1. é
åã®æåã® `[` ã¯å¯Ÿå¿ãã `]` ã§éããããŸãã
2. 次㫠`{` ã¯å¯Ÿå¿ãã `}` ã§éããããŸãã
3. `(` 㯠`)` ã§éããããŸãã
4. æåŸã« `]` ã®çŽåã« `{` ãè¿œå ãããå¿
èŠãããããã®çŽåŸã« `}` ãè¿œå ãããŸãã
ããããèæ
®ãããšãäžããããé
åãå®æããã圢ã¯æ¬¡ã®ããã«ãªããŸãã
```plaintext
[ { ( ) } { } ]
```
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(5 ÷ 2) + 10 - 3
ãã®æ°åŒã¯ã以äžã®æé ã§èšç®ãããŸãïŒ
1. 5ã2ã§å²ã (5 ÷ 2)
2. ãã®çµæã«10ã足ã ((5 ÷ 2) + 10)
3. ãã®å€ãã3ãåŒã ((5 ÷ 2) + 10 - 3)
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### ã¹ããã
1. 2ã€ã®æåå `s1` ãš `s2` ãæã€ã
2. 2次å
ã®é
å `dp` ãäœæãã`dp[i][j]` 㯠`s1` ã®æåã® `i` æåãš `s2` ã®æåã® `j` æåã®æé·å
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3. `dp` é
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4. `dp` é
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5. æçµçãªçµæ㯠`dp[len(s1)][len(s2)]` ã«å«ãŸããã
### Python ã³ãŒãäŸ
```python
def longest_common_subsequence(s1, s2):
# s1 ãš s2 ã®é·ããååŸ
len_s1 = len(s1)
len_s2 = len(s2)
# 2次å
é
å dp ãåæå
dp = [[0] * (len_s2 + 1) for _ in range(len_s1 + 1)]
# dp é
åãæŽæ°
for i in range(1, len_s1 + 1):
for j in range(1, len_s2 + 1):
if s1[i - 1] == s2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
else:
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
# æçµçãªçµæãè¿ã
return dp[len_s1][len_s2]
# ãã¹ã
s1 = "ããªãªkirin"
s2 = "ranrankiris"
result = longest_common_subsequence(s1, s2)
print(f"é£ç¶ããæé·å
±ééšåæååã®é·ã: {result}")
```
### å®è¡çµæ
ãã®ã³ãŒããå®è¡ãããšã`ããªãªkirin` ãš `ranrankiris` ã®é£ç¶ããæé·å
±ééšåæååã®é·ãã衚瀺ãããŸããå®éã®çµæ㯠`6` ã§ãïŒ`kirin`ïŒã
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