Update 053. Maximum Subarray

Author: haibarawu

053. Maximum Subarray

@@ -1,57 +1,44 @@
-/**
-Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
-For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
-the contiguous subarray [4,-1,2,1] has the largest sum = 6.
-**/
-
-//Java solution:
-
-public class Solution {
+/*
+ * @lc app=leetcode id=53 lang=java
+ *
+ * [53] Maximum Subarray
+ *
+ * https://leetcode.com/problems/maximum-subarray/description/
+ *
+ * algorithms
+ * Easy (42.32%)
+ * Total Accepted:    440.3K
+ * Total Submissions: 1M
+ * Testcase Example:  '[-2,1,-3,4,-1,2,1,-5,4]'
+ *
+ * Given an integer array nums, find the contiguous subarray (containing at
+ * least one number) which has the largest sum and return its sum.
+ * 
+ * Example:
+ * Input: [-2,1,-3,4,-1,2,1,-5,4],
+ * Output: 6
+ * Explanation: [4,-1,2,1] has the largest sum = 6.
+ * 
+ * Follow up:
+ * If you have figured out the O(n) solution, try coding another solution using
+ * the divide and conquer approach, which is more subtle.
+ * 
+ */
+ 
+class Solution {
     public int maxSubArray(int[] nums) {
-        if(nums == null || nums.length == 0){
+        if (nums.length == 0) {
             return 0;
         }
 
-        int sum;
-        int max;
-        int result = Integer.MIN_VALUE;
-        
-        for(int i = 0; i < nums.length; i++){
-            sum = 0;
-            max = Integer.MIN_VALUE;
-            
-            for(int j = i; j < nums.length; j++){
-                sum = sum + nums[j];
-                max = Math.max(sum, max);
-            }
-            result = Math.max(result, max);
-            //sum = 0;
-            //max = Integer.MIN_VALUE;
-        }
-        
-        return result;
-    }
-}
+        int newsum = nums[0];
+        int max = nums[0];
 
-
-/*
-public class Solution {
-    public int maxSubArray(int[] A) {
-        if (A == null || A.length == 0){
-            return 0;
+        for(int i = 1; i < nums.length; i++) {
+            newsum = Math.max(newsum + nums[i], nums[i]);
+            max = Math.max(max, newsum);
         }
 
-        int max = Integer.MIN_VALUE;
-        int sum = 0;
-        
-        for (int i = 0; i < A.length; i++) {
-            sum += A[i];
-            max = Math.max(max, sum);
-            sum = Math.max(sum, 0);
-        }
-
         return max;
     }
 }
-*/
-