LCR 017. 最小覆盖子串
解题思路
- need 记录满足条件的实际有效字符
- window记录当前窗口内的所有字符
- 遍历s 针对每一个字符送入window 计算valid 满足条件的有效字符个数
- 判断valid和need的尺寸 判断窗口是否需要收缩
- 如果需要收缩,计算窗口的长度,更新start 和 minLen
- 最后 如果左边字符在window中 移除该字符
class Solution {
public String minWindow(String s, String t) {
Map<Character, Integer> need = new HashMap<>();
Map<Character, Integer> window = new HashMap<>();
for (int i = 0; i < t.length(); i++) {
need.put(t.charAt(i), need.getOrDefault(t.charAt(i), 0) + 1);
}
int left = 0;
int right = 0;
int valid = 0;
int start = 0;
int minLength = Integer.MAX_VALUE;
while (right < s.length()) {
char c = s.charAt(right);
right++;
if (need.containsKey(c)) {
window.put(c, window.getOrDefault(c, 0) + 1);
if (window.get(c).equals(need.get(c))) {
valid++;
}
}
while (valid == need.size()) {
if (right - left < minLength) {
start = left;
minLength = right - left;
}
char d = s.charAt(left);
left++;
if (need.containsKey(d)) {
if (window.get(d).equals(need.get(d))) {
valid--;
}
window.put(d, window.get(d) - 1);
}
}
}
if (minLength == Integer.MAX_VALUE) {
return "";
} else {
return s.substring(start, start + minLength);
}
}
}