LeetCode Brushing Notes _53. Maximum Subsequence Sum

The topic is from LeetCode

53. Maximum Subsequence Sum

Other solutions or source code can be accessed: tongji4m3

description

Given an integer array nums, find a continuous sub-array with the largest sum (the sub-array contains at least one element), and return the largest sum.

Example:

输入: [-2,1,-3,4,-1,2,1,-5,4]
输出: 6
解释: 连续子数组 [4,-1,2,1] 的和最大，为 6。

Advanced:

If you have already implemented a solution with O(n) complexity, try to use a more subtle divide-and-conquer solution.

Ideas

Traverse from left to right, if it is greater than 0, keep it, otherwise don't. result is judged every moment

detail

Because the sub-array contains at least one element, the initialization cannot be 0, but the minimum value of int

Code

//子数组最少包含一个元素
public int maxSubArray(int[] nums)
{
    
    
    int N = nums.length;
    int result = Integer.MIN_VALUE,temp=0;//look,因为条件是至少包含
    for (int i = 0; i < N; i++)
    {
    
    
        temp = temp > 0 ? temp + nums[i] : nums[i];
        result = Math.max(result, temp);
    }
    return result;
}

Complexity analysis

time complexity

$O\left(N\right)$

Space complexity

$O\left(1\right)$