关于力扣:力扣-53-最大子序和

25次阅读

共计 1842 个字符,预计需要花费 5 分钟才能阅读完成。

最大子序和

1. 暴力法

工夫复杂度: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,顺次加上前面的数字,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. 动静布局

    工夫复杂度:O(N)
    空间复杂度:O(1)

  4. 一一加值比拟,存储最大值
  5. 一旦遇到加值后的后果 < 0,则只保留之前计算的最大值,从新开始下一个加值比拟

    class Solution
    {
    public:
     int maxSubArray(vector<int> &nums)
     {
         int result = INT_MIN;
         int numsSize = int(nums.size());
         //dp[i] 示意 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. 贪婪算法

    工夫复杂度:O(N)
    空间复杂度:O(1)
    与动静布局基本一致

    class Solution
    {
    public:
     int maxSubArray(vector<int> &nums)
     {
         // 相似寻找最大最小值的题目,初始值肯定要定义成实践上的最小最大值
         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,从新开始找子序串
             if (sum < 0)
             {sum = 0;}
         }
    
         return result;
     }
    };

    4. 分治法

    工夫复杂度:O(nlog(n))
    空间复杂度:O(log(n))

class Solution
{
public:
    int maxSubArray(vector<int> &nums)
    {
        // 相似寻找最大最小值的题目,初始值肯定要定义成实践上的最小最大值
        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);
        // 留神这里应是 mid + 1,否则 left + 1 = right 时,会有限循环
        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;
        // 留神这里 i = mid + 1,防止反复用到 nums[i]
        for (int i = mid + 1; i <= right; i++)
        {sum += nums[i];
            rightSum = max(rightSum, sum);
        }
        return (leftSum + rightSum);
    }
};

正文完
 0