MaximumSubarray-LeetCode

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#LeetCode #MaximumSubarray

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本文深入探讨了求解最大子数组和问题的高效算法，通过动态规划方法，实现了O(n)的时间复杂度。文章提供了详细的代码实现，包括使用递增变量curSum更新最大和的过程，以及使用dp变量进行状态转移的优化方案。

problem

Maximum Subarray

Easy

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.

key

我们定义一个和为第一位数，然后用curSum来保存递增量

ps ans–>answer cur–>cursor

solution

class Solution {
    public int maxSubArray(int[] nums) {
        int ans=nums[0], curSum=0;
        
        for (int i=0; i<nums.length; i++) {
            curSum = curSum + nums[i];
            ans = Math.max(ans, curSum);
            curSum = Math.max(0, curSum);
        }
        
        return ans;
    }
}

perfect

class Solution {
    public int maxSubArray(int[] nums) {
        int dp = nums[0], maxSum=nums[0];
        for (int i=1; i<nums.length; i++) {
            dp = dp<0?nums[i]:nums[i]+dp;
            maxSum=Math.max(maxSum, dp);
        }
        return maxSum;
    }
}