今天讲一个非常经典的滑动窗口的问题

这道题的意思很明显:

给你两个字符串s1与s2,判断s2中是否存在一个子串:它包含s1中所有字符且不包含其他字符

让我们先来写一下滑动窗口的模板:

    /*** 滑动窗口模板 * @param s1 * @param s2 */public static void model (String s1, String s2) {//用HashMap充当窗口Map<Character, Integer> need = new HashMap<>();Map<Character, Integer> window = new HashMap<>();for(char c : s1.toCharArray()){need.put(c, need.getOrDefault(c, 0) + 1);}//将s1放入需要的目标的值int left = 0, right = 0;//左右窗口int valid = 0;//满足目标的值while (right < s2.length()) {//left<right可写可不写,恒为truechar c = s2.charAt(right);//拿到rightright++;//判断右边进来的字母//相关代码//对左边界进行操作while (right - left >= s1.length()) {//相关代码}}return;}

接下来让我们一起做一下这道题

public static boolean checkInclusion (String s1, String s2) {Map<Character, Integer> need = new HashMap<>();Map<Character, Integer> window = new HashMap<>();for (int i = 0; i < s1.length(); i++) {need.put(s1.charAt(i), need.getOrDefault(s1.charAt(i), 0) + 1);}int left = 0, right = 0;int valid = 0;while (right < s2.length()) {char c = s2.charAt(right);right++;//右移右边界//判断右边进来的字母if (need.containsKey(c)) {window.put(c, window.getOrDefault(c, 0) + 1);//将字符c放入windowif (window.get(c).equals(need.get(c))) {valid++;//如果符合条件(window中该字符与need中相同)}}//对左边界进行操作while (right - left >= s1.length()) {if (valid == need.size()) {
//valid==need.size()的时候,right - left = s1.length()恒成立,无需多余判断return true;}char d = s2.charAt(left);left++;//右移左边界if (need.containsKey(d)) {//将字符d从window中删除if (need.get(d) == window.get(d)) {valid--;//如果删除之前该元素在window中与need相同}window.put(d, window.get(d) - 1);}}}return false;
}

其中需要注意的点我都已经做了注解,如果还有什么地方不明白,可以私信我