【力扣】 - 53. 最大子序和

最大子序和

1. 暴力法

时间复杂度：O(N^2)
空间复杂度：O(1)

1. 设置两层for循环
2. 存储第一个数字的值，依次加上后面的数字，只存储最大值
3. 依此类推

class Solution
{
public:
    int maxSubArray(vector<int> &nums)
    {
        int max = INT_MIN;
        int numsSize = int(nums.size());
        for (int i = 0; i < numsSize; i++)
        {
            int sum = 0;
            for (int j = i; j < numsSize; j++)
            {
                sum += nums[j];
                if (sum > max)
                {
                    max = sum;
                }
            }
        }
        return max;
    }
};

2. 动态规划

时间复杂度：O(N)
空间复杂度：O(1)

1. 逐个加值比较，存储最大值
2. 一旦遇到加值后的结果 < 0，则只保留之前计算的最大值，重新开始下一个加值比较

class Solution
{
public:
    int maxSubArray(vector<int> &nums)
    {
        int result = INT_MIN;
        int numsSize = int(nums.size());
        //dp[i]表示nums中以nums[i]结尾的最大子序和
        vector<int> dp(numsSize);
        dp[0] = nums[0];
        result = dp[0];
        for (int i = 1; i < numsSize; i++)
        {
            dp[i] = max(dp[i - 1] + nums[i], nums[i]);
            result = max(result, dp[i]);
        }
        return result;
    }
};

3.贪心算法

时间复杂度：O(N)
空间复杂度：O(1)
与动态规划基本一致

class Solution
{
public:
    int maxSubArray(vector<int> &nums)
    {
        //类似寻找最大最小值的题目，初始值一定要定义成理论上的最小最大值
        int result = INT_MIN;
        int numsSize = int(nums.size());
        int sum = 0;
        for (int i = 0; i < numsSize; i++)
        {
            sum += nums[i];
            result = max(result, sum);
            //如果sum < 0，重新开始找子序串
            if (sum < 0)
            {
                sum = 0;
            }
        }

        return result;
    }
};

4. 分治法

时间复杂度：O(nlog(n))
空间复杂度：O(log(n))

class Solution
{
public:
    int maxSubArray(vector<int> &nums)
    {
        //类似寻找最大最小值的题目，初始值一定要定义成理论上的最小最大值
        int result = INT_MIN;
        int numsSize = int(nums.size());
        result = maxSubArrayHelper(nums, 0, numsSize - 1);
        return result;
    }

    int maxSubArrayHelper(vector<int> &nums, int left, int right)
    {
        if (left == right)
        {
            return nums[left];
        }
        int mid = (left + right) / 2;
        int leftSum = maxSubArrayHelper(nums, left, mid);
        //注意这里应是mid + 1，否则left + 1 = right时，会无限循环
        int rightSum = maxSubArrayHelper(nums, mid + 1, right);
        int midSum = findMaxCrossingSubarray(nums, left, mid, right);
        int result = max(leftSum, rightSum);
        result = max(result, midSum);
        return result;
    }

    int findMaxCrossingSubarray(vector<int> &nums, int left, int mid, int right)
    {
        int leftSum = INT_MIN;
        int sum = 0;
        for (int i = mid; i >= left; i--)
        {
            sum += nums[i];
            leftSum = max(leftSum, sum);
        }

        int rightSum = INT_MIN;
        sum = 0;
        //注意这里i = mid + 1，避免重复用到nums[i]
        for (int i = mid + 1; i <= right; i++)
        {
            sum += nums[i];
            rightSum = max(rightSum, sum);
        }
        return (leftSum + rightSum);
    }
};