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feat: add solutions to lcof2 problem: No.017
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‎lcof2/剑指 Offer II 017. 含有所有字符的最短字符串/README.md

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Original file line numberDiff line numberDiff line change
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<pre>
2020
<strong>输入：</strong>s = &quot;ADOBECODEBANC&quot;, t = &quot;ABC&quot;
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<strong>输出：</strong>&quot;BANC&quot;
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<strong>输出：</strong>&quot;BANC&quot;
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<strong>解释：</strong>最短子字符串 &quot;BANC&quot; 包含了字符串 t 的所有字符 &#39;A&#39;、&#39;B&#39;、&#39;C&#39;</pre>
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<p><strong>示例 2：</strong></p>
@@ -57,22 +57,261 @@
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<!-- 这里可写通用的实现逻辑 -->
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滑动窗口，当窗口包含全部需要的的字符后，进行收缩，以求得最小长度
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进阶解法：利用 `count` 变量避免重复对 `need` 和 `window` 进行扫描
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minWindow(self, s: str, t: str) -> str:
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m, n = len(s), len(t)
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if n > m:
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return ""
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need, window = defaultdict(int), defaultdict(int)
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for c in t:
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need[c] += 1
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start, minLen = 0, float('inf')
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left, right = 0, 0
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while right < m:
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window[s[right]] += 1
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right += 1
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while self.check(need, window):
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if right - left < minLen:
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minLen = right - left
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start = left
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window[s[left]] -= 1
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left += 1
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return "" if minLen == float('inf') else s[start:start + minLen]
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def check(self, need, window):
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for k, v in need.items():
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if window[k] < v:
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return False
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return True
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```
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进阶解法
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```python
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class Solution:
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def minWindow(self, s: str, t: str) -> str:
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m, n = len(s), len(t)
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if n > m:
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return ""
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need, window = defaultdict(int), defaultdict(int)
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needCount, windowCount = 0, 0
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for c in t:
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if need[c] == 0:
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needCount += 1
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need[c] += 1
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start, minLen = 0, float('inf')
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left, right = 0, 0
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while right < m:
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ch = s[right]
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right += 1
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if ch in need:
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window[ch] += 1
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if window[ch] == need[ch]:
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windowCount += 1
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while windowCount == needCount:
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if right - left < minLen:
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minLen = right - left
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start = left
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ch = s[left]
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left += 1
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if ch in need:
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if window[ch] == need[ch]:
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windowCount -= 1
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window[ch] -= 1
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return "" if minLen == float('inf') else s[start:start + minLen]
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String minWindow(String s, String t) {
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int m = s.length(), n = t.length();
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if (n > m) {
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return "";
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}
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Map<Character, Integer> need = new HashMap<>();
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Map<Character, Integer> window = new HashMap<>();
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for (char ch : t.toCharArray()) {
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need.merge(ch, 1, Integer::sum);
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}
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int start = 0, minLen = Integer.MAX_VALUE;
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int left = 0, right = 0;
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while (right < m) {
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window.merge(s.charAt(right++), 1, Integer::sum);
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while (check(need, window)) {
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if (right - left < minLen) {
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minLen = right - left;
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start = left;
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}
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window.merge(s.charAt(left++), -1, Integer::sum);
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}
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}
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return minLen == Integer.MAX_VALUE ? "" : s.substring(start, start + minLen);
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}
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private boolean check(Map<Character, Integer> need, Map<Character, Integer> window) {
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for (Map.Entry<Character, Integer> entry : need.entrySet()) {
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if (window.getOrDefault(entry.getKey(), 0) < entry.getValue()) {
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return false;
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}
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}
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return true;
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}
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}
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```
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进阶解法
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```java
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class Solution {
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public String minWindow(String s, String t) {
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int m = s.length(), n = t.length();
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if (n > m) {
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return "";
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}
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Map<Character, Integer> need = new HashMap<>();
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Map<Character, Integer> window = new HashMap<>();
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int needCount = 0, windowCount = 0;
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for (char ch : t.toCharArray()) {
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if (!need.containsKey(ch)) {
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needCount++;
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}
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need.merge(ch, 1, Integer::sum);
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}
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int start = 0, minLen = Integer.MAX_VALUE;
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int left = 0, right = 0;
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while (right < m) {
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char ch = s.charAt(right++);
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if (need.containsKey(ch)) {
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int val = window.getOrDefault(ch, 0) + 1;
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if (val == need.get(ch)) {
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windowCount++;
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}
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window.put(ch, val);
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}
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while (windowCount == needCount) {
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if (right - left < minLen) {
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minLen = right - left;
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start = left;
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}
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ch = s.charAt(left++);
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if (need.containsKey(ch)) {
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int val = window.get(ch);
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if (val == need.get(ch)) {
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windowCount--;
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}
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window.put(ch, val - 1);
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}
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}
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}
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return minLen == Integer.MAX_VALUE ? "" : s.substring(start, start + minLen);
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}
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}
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```
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### **Go**
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```go
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func minWindow(s string, t string) string {
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m, n := len(s), len(t)
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if n > m {
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return ""
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}
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need, window := make(map[byte]int), make(map[byte]int)
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for _, r := range t {
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need[byte(r)]++
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}
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start, minLen := 0, math.MaxInt32
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left, right := 0, 0
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for right < m {
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window[s[right]]++
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right++
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for check(need, window) {
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if right-left < minLen {
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minLen = right - left
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start = left
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}
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window[s[left]]--
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left++
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}
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}
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if minLen == math.MaxInt32 {
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return ""
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}
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return s[start : start+minLen]
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}
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func check(need, window map[byte]int) bool {
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for k, v := range need {
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if window[k] < v {
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return false
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}
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}
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return true
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}
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```
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进阶解法
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```go
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func minWindow(s string, t string) string {
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m, n := len(s), len(t)
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if n > m {
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return ""
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}
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need, window := make(map[byte]int), make(map[byte]int)
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needCount, windowCount := 0, 0
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for _, r := range t {
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if need[byte(r)] == 0 {
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needCount++
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}
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need[byte(r)]++
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}
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start, minLen := 0, math.MaxInt32
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left, right := 0, 0
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for right < m {
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ch := s[right]
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right++
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if v, ok := need[ch]; ok {
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window[ch]++
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if window[ch] == v {
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windowCount++
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}
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}
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for windowCount == needCount {
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if right-left < minLen {
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minLen = right - left
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start = left
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}
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ch = s[left]
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left++
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if v, ok := need[ch]; ok {
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if window[ch] == v {
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windowCount--
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}
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window[ch]--
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}
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}
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}
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if minLen == math.MaxInt32 {
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return ""
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}
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return s[start : start+minLen]
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}
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```
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### **...**

‎lcof2/剑指 Offer II 017. 含有所有字符的最短字符串/Solution.go

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// func minWindow(s string, t string) string {
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// m, n := len(s), len(t)
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// if n > m {
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// return ""
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// }
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// need, window := make(map[byte]int), make(map[byte]int)
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// for _, r := range t {
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// need[byte(r)]++
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// }
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// start, minLen := 0, math.MaxInt32
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// left, right := 0, 0
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// for right < m {
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// window[s[right]]++
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// right++
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// for check(need, window) {
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// if right-left < minLen {
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// minLen = right - left
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// start = left
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// }
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// window[s[left]]--
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// left++
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// }
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// }
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// if minLen == math.MaxInt32 {
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// return ""
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// }
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// return s[start : start+minLen]
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// }
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// func check(need, window map[byte]int) bool {
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// for k, v := range need {
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// if window[k] < v {
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// return false
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// }
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// }
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// return true
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// }
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func minWindow(s string, t string) string {
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m, n := len(s), len(t)
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if n > m {
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return ""
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}
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need, window := make(map[byte]int), make(map[byte]int)
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needCount, windowCount := 0, 0
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for _, r := range t {
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if need[byte(r)] == 0 {
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needCount++
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}
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need[byte(r)]++
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}
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start, minLen := 0, math.MaxInt32
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left, right := 0, 0
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for right < m {
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ch := s[right]
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right++
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if v, ok := need[ch]; ok {
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window[ch]++
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if window[ch] == v {
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windowCount++
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}
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}
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for windowCount == needCount {
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if right-left < minLen {
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minLen = right - left
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start = left
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}
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ch = s[left]
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left++
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if v, ok := need[ch]; ok {
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if window[ch] == v {
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windowCount--
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}
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window[ch]--
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}
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}
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}
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if minLen == math.MaxInt32 {
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return ""
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}
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return s[start : start+minLen]
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}

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