edit 238

fishercoder1534 · fishercoder1534 · commit d38ae97e5e9f · 2017-08-03T08:36:15.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_238.java b/src/main/java/com/fishercoder/solutions/_238.java
@@ -1,26 +1,37 @@
 package com.fishercoder.solutions;
 
 /**
- * Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
-
- Solve it without division and in O(n).
-
- For example, given [1,2,3,4], return [24,12,8,6].
+ * 238. Product of Array Except Self
+ *
+ * Given an array of n integers where n > 1, nums,
+ * return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
+ * Solve it without division and in O(n).
+ * For example, given [1,2,3,4], return [24,12,8,6].
 
  Follow up:
- Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)
+ Could you solve it with constant space complexity?
+ (Note: The output array does not count as extra space for the purpose of space complexity analysis.)
  */
+
 public class _238 {
 
+    /**Very straightforward idea: iterate through the array twice:
+     * first time: get res[i] = res[i-1]*nums[i-1]
+     * second time: have a variable called right, which means all the numbers product to its right, then do
+     * res[i] *= right;
+     * right *= nums[i];
+     * that's it.
+
+     * This could be very well illustrated with this example: [1,2,3,4]*/
     public int[] productExceptSelf(int[] nums) {
         int n = nums.length;
         int[] result = new int[n];
         result[0] = 1;
         for (int i = 1; i < n; i++) {
-            result[i] = result[i-1]*nums[i-1];
+            result[i] = result[i - 1] * nums[i - 1];
         }
         int right = 1;
-        for (int i = n-1; i >= 0; i--) {
+        for (int i = n - 1; i >= 0; i--) {
             result[i] *= right;
             right *= nums[i];
         }
