实现内容：
假如有这样一个数组，A[] = {13,-2,-25,20,-3,-16,-23,18,20,-7,12,-5,-22,15,-4,7};要求得到一个总值最大的子数组。例如A[]的一个最大子数组为A[7..10]={18,20,-7,12}。
求解思想:
1）暴力求解
循环每一个元素，并从每一个元素的下标开始累加，每次选取较大的和。例如从第一个元素开始，第一次累加是13，那么记录最大的maxSubArray为13；第二次累加为13-2=11，比13小，maxSubArray还是为13；然后继续累加，选择最大的maxSubArray;最后循环下一个元素。A为输入数组，n为最大子数组起始元素下标,m为最大子数组终止元素下标,maxSum为最大子数组的和
伪代码：
FindMaxSubArray(A)
maxSum=0
for i = 1 to A.length
let curSum = 0
for j = i to A.length
curSum = curSum + A[j]
if curSum > maxSum
maxSum = curSum
let n = i,m = j
return n,m,maxSum
2）分治思想
分治思想，把一个问题分解为很多小问题，先解决小问题然后再逐级向上解决整个问题。该问题的解决思想跟归并排序类似，先把数组拆分为两个子序列，然后两个子序列继续拆分。同样也需要一个辅助函数，用于计算子序列数组对应下标的最大子数组和。
伪代码：
FindMaxCrossingSubArray(A,low,mid,high) //辅助函数
left-sum = -∞
sum = 0
for i = mid down to low
sum = sum + A[i]
if sum > left-sum
left-sum = sum
max-left = i
right-sum = -∞
sum = 0
for j = mid + 1 to high
sum = sum + A[j]
if sum > righ-sum
right-sum = sum
max-right = j
return (max-left,max-right,left-sum+right-sum)
FindMaxSubArrau(A, low, high) //主函数
if high == low
return (low,high,A[low]) //分解为只有一个元素时
else
mid = (low+high)/2
(left-low,left-high,left-sum) = FindMaxSubArrau(A,low,mid)
(right-low,right-high,right-sum) = FindMaxSubArrau(A,mid+1,high)
(cross-low,cross-high,cross-sum) = FindMaxCrossingSubArray(A,low,mid,high)
if left-sum >= right-sum and left-sum >= cross-sum
return (left-low,left-high,left-sum)
else if right-sum >= left-sum and right-sum >= cross-sum
return (right-low,right-high,right-sum)
else
return (cross-low,cross-high,cross-sum)
3）最优解
假如有这样一个数组，数组元素大于2并且数组元素里面有正负，那么最大子数组起始元素一定是正整数，每次从正整数开始计算，求出最大的子数组。i为最大子数组起始元素下标，j为最大子数组终止元素下标,maxSum为最大字数组的和
伪代码：
FindMaxSubArray(A) //注释的代码功能为记录最大子数组起始元素和终止元素的下标
flag = maxSum = curSum = 0
//i = j = 0;
for m = 0 to A.length
curSum = curSum + A[m]
if curSum > maxSum
//if flag != i
// i = flag
//j = m
maxSum = curSum
if curSum < 0
curSum = 0
//if m < A.length-1
// if A[m+1] > 0
// flag = m + 1
//return i,j,maxSum
return maxSum
class FindMaxSubArray{
public static void main(String[] args) {
int[] A= {13,-2,-25,20,-3,-16,-23,18,20,-7,12,-5,-22,15,-4,7};
//int[] A = {13,-2,-2,10};
//存放最大字数组的起始下标，终止下标，数组和
int[] result = new int[3];
//result = FindMaxSubArray1(A,result);
//result = FindMaxSubArray2(A,result,0,A.length-1);
result = FindMaxSubArray3(A,result);
for (int i: result) {
System.out.print(i+" ");
}
}
//暴力求解
public static int[] FindMaxSubArray1(int[] A,int[] result){
int i = 0, j = 0;
int maxSum = 0;
int len = A.length;
for (int m = 0;m < len; m++) {
int curSum = 0;
for (int n = m; n < len; n++) {
curSum += A[n];
if (curSum > maxSum){
maxSum = curSum;
i = m;
j = n;
}
}
}
getResult(result,i,j,maxSum);
return result;
}
//分治思想
public static int[] FindMaxSubArray2(int[] A,int[] result,int low,int high){
if(low==high){
result = getResult(result,low,high,A[low]);
return result;
}else{
int mid = (low+high)/2;
int[] leftResult = new int[3];
int[] rightResult = new int[3];
int[] crossResult = new int[3];
leftResult = FindMaxSubArray2(A,leftResult,low,mid);
rightResult = FindMaxSubArray2(A,rightResult,mid+1,high);
crossResult = FindMaxCrossingSubArray(A,crossResult,low,mid,high);
if(leftResult[2]>=rightResult[2]&&leftResult[2]>=crossResult[2]){
result = leftResult;
// System.out.print("left: ");
// for(int i:result){
// System.out.print(i+" ");
// }
}
else if (rightResult[2]>=leftResult[2]&&rightResult[2]>=crossResult[2]) {
result = rightResult;
//System.out.print("right: ");
// for(int i:result){
// System.out.print(i+" ");
// }
}else{
result = crossResult;
// System.out.print("cross: ");
// for(int i:result){
// System.out.print(i+" ");
// }
}
//System.out.println();
return result;
}
}
//分治过程中用到的辅助函数
public static int[] FindMaxCrossingSubArray(int[] A,int[] result,int low,int mid,int high){
int leftSum = A[mid];
int curLeftSum = 0;
int left = mid;
for(int i = mid; i >= low; i--){
curLeftSum += A[i];
if(curLeftSum > leftSum){
leftSum = curLeftSum;
left = i;
}
}
int rightSum = A[mid+1];
int curRightSum = 0;
int right = mid+1;
for(int i = mid+1; i <= high; i++){
curRightSum += A[i];
if(curRightSum > rightSum){
rightSum = curRightSum;
right = i;
}
}
result = getResult(result,left,right,leftSum+rightSum);
return result;
}
//最优解
public static int[] FindMaxSubArray3(int[] A,int[] result){
int m;
int i = 0;//记录起始元素的下标
int j = 0;//记录终止元素的下标
/*//记录 上一次maxSum最大时终止元素的下标
int k = 0;*/
int flag = 0;//记录curSum又一次从负整数变为正整数时起始元素的下标
int maxSum = 0;//最大子数组和
int curSum = 0;//当前最大子数组和
for (m=0;m<A.length ;m++ ) {
curSum += A[m];
if(curSum>maxSum){
//当前最大子数组和较大时，更换起始元素下标
if (flag != i) {
i = flag;
}
j = m;//更换当前终止元素下标
maxSum = curSum;
//更换上一次最大子数组终止元素的下标
/*if(k!=0||k!=j){
k++;
}
if ((j-k)!=1||j==0){
k = m;
}*/
}
if(curSum<0){
curSum = 0;
//记录下一个正整数为起始元素的下标
if (m<(A.length-1)) {
if (A[m+1]>0) {
flag = m+1;
}
}
}
}
result = getResult(result,i,j,maxSum);
return result;
}
public static int[] getResult(int[] result,int i,int j,int maxSum){
result[0] = i;
result[1] = j;
result[2] = maxSum;
return result;
}
}