diff --git a/solution/0400-0499/0400.Nth Digit/Solution.java b/solution/0400-0499/0400.Nth Digit/Solution.java
new file mode 100644
index 0000000000000..e0062db95b76a
--- /dev/null
+++ b/solution/0400-0499/0400.Nth Digit/Solution.java	
@@ -0,0 +1,29 @@
+class Solution {
+    /***
+     * 12345678910111213
+     * 规律个位数9个数一共有9*1,两位数90个数 一共有90*2个数字,三位数有900个数一共有900*3个数字,以此类推
+     * 举例15,15-9=6,6/2=3...0,余数是0,那么这个数值value=10*(2-1)+(3-1)=12,整除取最后一位  12%10=2
+     * 举例14,14-9=5,5/2=2...1,余数不为0,那么这个数值value=10*(2-1)+2=12,则为这个数的第余数个 12/(10*(2-1))%10=1
+     */
+    public int findNthDigit(int n) {
+        long max = 9;
+        long num = n;
+        long digits = 1;
+        while (n > 0) {
+            if (num - max * digits > 0) {
+                num = num - max * digits;
+                digits++;
+                max = max * 10;
+            } else {
+                long count = num / digits;
+                long childDigits = num % digits;
+                if (childDigits == 0) {
+                    return (int) (((long) Math.pow(10, digits - 1) + count - 1) % 10);
+                } else {
+                    return (int) (((long) Math.pow(10, digits - 1) + count) / ((long) Math.pow(10, (digits - childDigits))) % 10);
+                }
+            }
+        }
+        return 0;
+    }
+}
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