01-复杂度2 Maximum Subsequence Sum（25 分）
Given a sequence of K integers { N
1 , N2 , …, NK
}. A continuous subsequence is defined to be { N
i
, N
i+1
, …, N
j
} where 1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.
Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.
Input Specification:
Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (≤10000). The second line contains K numbers, separated by a space.
Output Specification:
For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.
Sample Input:
10
-10 1 2 3 4 -5 -23 3 7 -21
Sample Output:
10 1 4
/**
* 01-复杂度2 Maximum Subsequence Sum（25 分）
* 求解最大子列和与最大子列和的位置，不对啊，部分正确，部分错误
* @param inputArray
* @return
*/
public static void Maximum_Subsequence_Sum(int[] inputArray){
int maxSum = 0; //子列最大和
int thisSum = 0; //当前可能子列最大和
int Lindex = 0; //左边索引
int Rindex = 0; // 右边索引
int maybe_Lindex=0;//可能成为最大子列和的第一个结点
int flag_max=0;//flag represent whether max has been rewrite
int flag = 0; //thisSum归零一次，有可能后续子数列会出现大于当前max的数列
for (int i = 0; i < inputArray.length; ++i){
thisSum +=inputArray[i]; //向右累加
if (thisSum > maxSum){
flag_max = 1;
if (flag == 0){
Lindex = maybe_Lindex;
flag =1;
}
Rindex = i; //更新右边的索引
maxSum = thisSum; //发现更大的则更新当前的结果
}
else if (thisSum < 0){ //如果当前子列和小于0，抛弃
maybe_Lindex = i+1;
thisSum = 0;
flag = 0;
}
}
if (flag_max == 0){
System.out.println(0 + " " + inputArray[0] + " " + (inputArray[inputArray.length-1]));
}
else{
System.out.println(maxSum + " " +inputArray[Lindex] + " " + inputArray[Rindex]);
}
}