add 0520,0521,0522 solution for java

solution/0500-0599/0520.Detect Capital/Solution.java

@@ -0,0 +1,24 @@
+class Solution {
+    public boolean detectCapitalUse(String word) {
+        char[] cs = word.toCharArray();
+        int upper = 0;
+        int lower = 0;
+        for (int i = 0; i < cs.length; i++) {
+            if (cs[i] >= 'a') {
+                lower++;
+            } else {
+                upper++;
+            }
+        }
+        if (upper == cs.length) {
+            return true;
+        }
+        if (lower == cs.length) {
+            return true;
+        }
+        if (upper == 1 && cs[0] < 'a') {
+            return true;
+        }
+        return false;
+    }
+}

solution/0500-0599/0521.Longest Uncommon Subsequence I/Solution.java

@@ -0,0 +1,7 @@
+class Solution {
+    public int findLUSlength(String a, String b) {
+        if (a.equals(b))
+            return -1;
+        return Math.max(a.length(), b.length());
+    }
+}

solution/0500-0599/0522.Longest Uncommon Subsequence II/Solution.java

@@ -0,0 +1,38 @@
+class Solution {
+    public int findLUSlength(String[] strs) {
+        int res = -1;
+        if (strs == null || strs.length == 0) {
+            return res;
+        }
+        if (strs.length == 1) {
+            return strs[0].length();
+        }
+        // 两两比较
+        // 1、存在子串,直接不比较后面的字符串
+        // 2、不存在子串,判断当前字符串是否是最长的字符串
+        for (int i = 0, j; i < strs.length; i++) {
+            for (j = 0; j < strs.length; j++) {
+                if (i == j) {
+                    continue;
+                }
+                // 根据题意，子串 可以 不是 原字符串中 连续的子字符串
+                if (isSubsequence(strs[i], strs[j])) {
+                    break;
+                }
+            }
+            if (j == strs.length) {
+                res = Math.max(res, strs[i].length());
+            }
+        }
+        return res;
+    }
+
+    public boolean isSubsequence(String x, String y) {
+        int j = 0;
+        for (int i = 0; i < y.length() && j < x.length(); i++) {
+            if (x.charAt(j) == y.charAt(i))
+                j++;
+        }
+        return j == x.length();
+    }
+}