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Commit 7a377bd

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add python of 53
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8 files changed

+60
-14
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  • codes
    • java/leetcodes/src/main/java/com/hit/basmath/interview/top_interview_questions/easy_collection/dynamic_programming
    • python/leetcodes/src/main/python/com/hit/basmath

8 files changed

+60
-14
lines changed

‎README.md‎

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‎codes/java/leetcodes/src/main/java/com/hit/basmath/interview/top_interview_questions/easy_collection/dynamic_programming/_53.java‎

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*/
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public class _53 {
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public int maxSubArray(int[] nums) {
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int maxSoFar = nums[0], maxEndingHere = nums[0];
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for (int i = 1; i < nums.length; ++i) {
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maxEndingHere = Math.max(maxEndingHere + nums[i], nums[i]);
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maxSoFar = Math.max(maxSoFar, maxEndingHere);
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int n = nums.length;
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// dp[i] means the maximum subarray ending with nums[i]
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int[] dp = new int[n];
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dp[0] = nums[0];
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int max = dp[0];
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for (int i = 1; i < n; i++) {
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dp[i] = nums[i] + (dp[i - 1] > 0 ? dp[i - 1] : 0);
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max = Math.max(max, dp[i]);
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}
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return maxSoFar;
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return max;
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}
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}
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"""
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53. Maximum Subarray
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Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
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Example:
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Input: [-2,1,-3,4,-1,2,1,-5,4],
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Output: 6
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Explanation: [4,-1,2,1] has the largest sum = 6.
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Follow up:
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If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
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"""
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def maxSubArray(self, nums):
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"""
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:type nums: List[int]
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:rtype: int
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"""
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numsLen = len(nums)
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dp = [nums[0]]
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maxSubSum = nums[0]
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i = 1
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while i < numsLen:
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if dp[i - 1] > 0:
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temp = dp[i - 1]
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else:
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temp = 0
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dp.append(nums[i] + temp)
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if maxSubSum < dp[i]:
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maxSubSum = dp[i]
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i += 1
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return maxSubSum

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