plusone

LRH1993 · LRH1993 · commit 06c40b1e6ef6 · 2017-09-18T09:51:06.000+08:00
diff --git a/SUMMARY.md b/SUMMARY.md
@@ -192,7 +192,8 @@
     - [Stone Game](/algorithm/LeetCode/Dynamic-Programming/Stone-Game.md)
   - [Array](/algorithm/LeetCode/Array.md)
     - [Partition Array](/algorithm/LeetCode/Array/Partition-Array.md)
-    - [Subarray Sum](/algorithm/LeetCode/Array/Subarray-Sum.md) 
+    - [Subarray Sum](/algorithm/LeetCode/Array/Subarray-Sum.md)
+    - [Plus One](/algorithm/LeetCode/Array/plus-one.md) 
   - [String](/algorithm/LeetCode/String.md)
     - [Restore IP Addresses](/algorithm/LeetCode/String/ip.md)
 
diff --git a/algorithm/LeetCode/Array/plus-one.md b/algorithm/LeetCode/Array/plus-one.md
@@ -0,0 +1,45 @@
+## 一、题目
+
+> Given a non-negative number represented as an array of digits, plus one to the number.
+>
+> The digits are stored such that the most significant digit is at the head of the list.
+>
+> Example
+>
+> Given [1,2,3] which represents 123, return [1,2,4].
+>
+> Given [9,9,9] which represents 999, return [1,0,0,0].
+
+给一个包含非负整数的数组，其中每个值代表该位数的值，对这个数加1。
+
+## 二、解题思路
+
+1. 数组的最后一个数是个位数，所以从后面开始读，个位数+1后，如果有进位，存储进位值，没有直接存储。
+2. 处理十位数，如果个位数有进位，十位数+1,在判断十位数有没有进位。
+3. 重复上面的动作直到没有进位。
+
+## 三、解题代码
+
+```java
+public class Solution {
+    
+    public int[] plusOne(int[] digits) {
+        int carries = 1;
+        for(int i = digits.length-1; i>=0 && carries > 0; i--){  // fast break when carries equals zero
+            int sum = digits[i] + carries;
+            digits[i] = sum % 10;
+            carries = sum / 10;
+        }
+        if(carries == 0)
+            return digits;
+            
+        int[] rst = new int[digits.length+1];
+        rst[0] = 1;
+        for(int i=1; i< rst.length; i++){
+            rst[i] = digits[i-1];
+        }
+        return rst;
+    }
+}
+```
+
