LEETCODE--Maximum Subarray

最新推荐文章于 2022-03-25 16:53:26 发布

prettysky123

最新推荐文章于 2022-03-25 16:53:26 发布

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本文介绍了一种寻找具有最大和的连续子数组的方法，并提供了两种不同的实现方式。第一种方法通过迭代更新当前值来确定最大子数组的和，第二种方法使用了累加和的方式并结合数学比较来获取结果。

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Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.
click to show more practice.
More practice:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
方法一：

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
      //  int p = 0;
        int q = 0;
        int max = nums[0];
        int val = nums[0];
        while( q < nums.size() - 1 ){
            q++;
            if( val < 0 )
                val = nums[q];
            else
                val += nums[q];
            if( max < val )
                max = val;
        }
        return max;
    }
};

方法二：

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int ls = nums[0];
        int val = 0;
        for(int i = 0; i < nums.size(); i++){
            val += nums[i];
            ls = max( val , ls );
            val = max( val , 0 );
        }
        return ls;
    }
};