int MaxSubSum1(const vector <int> & a) {
	int maxSum=0;

	for (int i=0; i<a.size(); i++) // i is the left end position of the subsequence
		for (int j=i; j<a.size(); j++) {// j is the position of the right end of the subsequence
			int thisSum=0; // thisSum is sum of subsequence from A [i] to A [j]

			for (int k=i; k<=j; k++)
				thisSum+=a[k];

			if (thisSum>maxSum)//If you just get this subsequence and bigger				maxSum=thisSum;//Update result		
}
	return maxSum;
}

（2）Algorithm 2

时间复杂度：O(N^2)

int MaxSubSum2(const vector <int> & a) {
	int maxSum=0;

	for (int i=0; i<a.size(); i++) {
		thisSum=0;
		for (int j=i; j<a.size(); j++) {
			thisSum+=a[j];// For the same i and different j, just accumulate 1 term on the basis of j-1 cycles

			if (thisSum>maxSum)
				maxSum=thisSum;
		}
	}
	return maxSum;
}

（3）Algorithm 3

时间复杂度：O(N)

int maxSumRec (const verctor<int> & a, int left, int right) {
    if (left==right)
        if (a[left]>0)
            return a[left];
        else
            return 0;

    int center = (left+right)/2;
    int maxLeftSum=maxSumRec(a, left, center);
    int maxRightSum=maxSumRec(a, center+1, right);

    int maxLeftBorderSum=0, leftBorderSum=0;
    for (int i=center; i>=left; i--) {
        leftBorderSum += a[i];
        if (leftBorderSum>maxLeftBorderSum)
            maxLeftBorderSum=leftBorderSum;
    }

    int maxRightBorderSum=0; rightBorderSum=0;
    for (int j=center+1; j<=right; j++) {
        rightBorderSum+=a[j];
        if (rightBorderSum>maxrightBorderSum)
            maxRightBorderSum=rightBorderSum;
    }

    return max3(maxLeftSum, maxRightSum, 
              maxLeftBorderSum+maxRightBorderSum);
}

int maxSubSum3 (const vector <int> & a) {
    return maxSumRec(a, 0, a.size()-1);
}

（4）Algorithm 4

时间复杂度：O(N)

int MaxSubSum4(const vector <int> & a) {
	int maxSum=0, thisSum=0;
	for (int j=0; j<a.size(); j++) {
		thisSum+=a[j];//Accumulate right
		if (thisSum>maxSum)//Find bigger and update current results			maxSum=thisSum;
		else if (thisSum<0)//If the current subsequence sum is negative			thisSum=0;//Then it ’s impossible to increase the sum of the latter part and discard it.	}
	return maxSum;
}