feat: add solutions to lc problem: No.0076

No.0076.Minimum Window Substring

Author: yanglbme

solution/0000-0099/0076.Minimum Window Substring/README.md

@@ -66,7 +66,29 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def minWindow(self, s: str, t: str) -> str:
+        ans = ''
+        m, n = len(s), len(t)
+        if m < n:
+            return ans
+        need = Counter(t)
+        window = Counter()
+        i, cnt, mi = 0, 0, float('inf')
+        for j, c in enumerate(s):
+            window[c] += 1
+            if need[c] >= window[c]:
+                cnt += 1
+            while cnt == n:
+                if j - i + 1 < mi:
+                    mi = j - i + 1
+                    ans = s[i: j + 1]
+                c = s[i]
+                if need[c] >= window[c]:
+                    cnt -= 1
+                window[c] -= 1
+                i += 1
+        return ans
 ```
 
 ### **Java**
@@ -202,4 +224,47 @@ public:
 };
 ```
 
+### **Go**
+
+```go
+func minWindow(s string, t string) string {
+	ans := ""
+	m, n := len(s), len(t)
+	if m < n {
+		return ans
+	}
+	need := make([]int, 128)
+	for _, c := range t {
+		need[c] += 1
+	}
+	window := make([]int, 128)
+	i, cnt, mi := 0, 0, m+1
+	for j, c := range s {
+		window[c]++
+		if need[c] >= window[c] {
+			cnt++
+		}
+		for cnt == n {
+			if j-i+1 < mi {
+				mi = j - i + 1
+				ans = s[i : j+1]
+			}
+			c = rune(s[i])
+			if need[c] >= window[c] {
+				cnt--
+			}
+			window[c]--
+			i++
+		}
+	}
+	return ans
+}
+```
+
+### **...**
+
+```
+
+```
+
 <!-- tabs:end -->

solution/0000-0099/0076.Minimum Window Substring/README_EN.md

@@ -56,7 +56,29 @@ Since the largest window of s only has one 'a', return empty string.
 ### **Python3**
 
 ```python
-
+class Solution:
+    def minWindow(self, s: str, t: str) -> str:
+        ans = ''
+        m, n = len(s), len(t)
+        if m < n:
+            return ans
+        need = Counter(t)
+        window = Counter()
+        i, cnt, mi = 0, 0, float('inf')
+        for j, c in enumerate(s):
+            window[c] += 1
+            if need[c] >= window[c]:
+                cnt += 1
+            while cnt == n:
+                if j - i + 1 < mi:
+                    mi = j - i + 1
+                    ans = s[i: j + 1]
+                c = s[i]
+                if need[c] >= window[c]:
+                    cnt -= 1
+                window[c] -= 1
+                i += 1
+        return ans
 ```
 
 ### **Java**
@@ -190,4 +212,47 @@ public:
 };
 ```
 
+### **Go**
+
+```go
+func minWindow(s string, t string) string {
+	ans := ""
+	m, n := len(s), len(t)
+	if m < n {
+		return ans
+	}
+	need := make([]int, 128)
+	for _, c := range t {
+		need[c] += 1
+	}
+	window := make([]int, 128)
+	i, cnt, mi := 0, 0, m+1
+	for j, c := range s {
+		window[c]++
+		if need[c] >= window[c] {
+			cnt++
+		}
+		for cnt == n {
+			if j-i+1 < mi {
+				mi = j - i + 1
+				ans = s[i : j+1]
+			}
+			c = rune(s[i])
+			if need[c] >= window[c] {
+				cnt--
+			}
+			window[c]--
+			i++
+		}
+	}
+	return ans
+}
+```
+
+### **...**
+
+```
+
+```
+
 <!-- tabs:end -->

solution/0000-0099/0076.Minimum Window Substring/Solution.go

@@ -0,0 +1,32 @@
+func minWindow(s string, t string) string {
+	ans := ""
+	m, n := len(s), len(t)
+	if m < n {
+		return ans
+	}
+	need := make([]int, 128)
+	for _, c := range t {
+		need[c] += 1
+	}
+	window := make([]int, 128)
+	i, cnt, mi := 0, 0, m+1
+	for j, c := range s {
+		window[c]++
+		if need[c] >= window[c] {
+			cnt++
+		}
+		for cnt == n {
+			if j-i+1 < mi {
+				mi = j - i + 1
+				ans = s[i : j+1]
+			}
+			c = rune(s[i])
+			if need[c] >= window[c] {
+				cnt--
+			}
+			window[c]--
+			i++
+		}
+	}
+	return ans
+}

solution/0000-0099/0076.Minimum Window Substring/Solution.py

@@ -0,0 +1,23 @@
+class Solution:
+    def minWindow(self, s: str, t: str) -> str:
+        ans = ''
+        m, n = len(s), len(t)
+        if m < n:
+            return ans
+        need = Counter(t)
+        window = Counter()
+        i, cnt, mi = 0, 0, float('inf')
+        for j, c in enumerate(s):
+            window[c] += 1
+            if need[c] >= window[c]:
+                cnt += 1
+            while cnt == n:
+                if j - i + 1 < mi:
+                    mi = j - i + 1
+                    ans = s[i: j + 1]
+                c = s[i]
+                if need[c] >= window[c]:
+                    cnt -= 1
+                window[c] -= 1
+                i += 1
+        return ans