🎉 #53. Maximum Subarray

Author: GoogTech

#53. Maximum Subarray/Solution.java

@@ -0,0 +1,25 @@
+/*
+ * @Author: Goog Tech
+ * @Date: 2020-09-15 13:44:15
+ * @LastEditTime: 2020-09-15 13:44:36
+ * @Description: https://leetcode-cn.com/problems/maximum-subarray/
+ * @FilePath: \leetcode-googtech\#53. Maximum Subarray\Solution.java
+ * @WebSite: https://algorithm.show/
+ */
+
+class Solution {
+    // DP : 动态规划,解题思路如下所示:
+    // 1. 首先对数组进行遍历,设当前最大连续子序列和为 sum, 结果为 result
+    // 2. 若 sum > 0,则说明 sum 对结果有增益效果,进而将当前所遍历的数字累加到 sum
+    // 3. 若 sum <= 0,则说明 sum 对结果无增益效果,进而舍弃当前所遍历数字,并将 sum 更新为当前所遍历的数字
+    // 4. 每次比较 sum 与 result 的大小,即将最大值置为 result,遍历结束后返回结果即可
+    public int maxSubArray(int[] nums) {
+        int result = nums[0], sum = 0;
+        for(int num : nums) {
+            if(sum > 0) sum += num;
+            else sum = num;
+            result = Math.max(result, sum);
+        }
+        return result;
+    }
+}

#53. Maximum Subarray/Solution.py

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+'''
+Author: Goog Tech
+Date: 2020-09-15 13:44:20
+LastEditTime: 2020-09-15 13:44:46
+Description: https://leetcode-cn.com/problems/maximum-subarray/
+FilePath: \leetcode-googtech\#53. Maximum Subarray\Solution.py
+WebSite: https://algorithm.show/
+'''
+
+class Solution(object):
+    # DP : 动态规划
+    def maxSubArray(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        for i in range(1, len(nums)):
+            if nums[i - 1] > 0: 
+                nums[i] += nums[i - 1]
+        return max(nums)