“最大子序列和”算法 java

public class Test {
public static void main(String[] args) {
int[] a = {-2, 11, -4, 13, -5, -2};//最大子序列和为20
int[] b = {-6, 2, 4, -7, 5, 3, 2, -1, 6, -9, 10, -2};//最大子序列和为16
System.out.println(maxSubSum4(a));
System.out.println(maxSubSum4(b));
}
//最大子序列求和算法一
public static int maxSubSum1(int[] a){

int maxSum = 0;

//从第i个开始找最大子序列和
for(int i = 0; i < a.length; i++) {

//找第i到j的最大子序列和
for(int j = i; j<a.length; j++) {

int thisSum = 0;

//计算从第i个开始，到第j个的和thisSum
for(int k = i; k<=j; k++){
thisSum += a[k];
}
//如果第i到第j个的和小于thisSum，则将thisSum赋值给maxSum
if(thisSum>maxSum) {
maxSum = thisSum;
}
}
}
return maxSum;
}

public static int maxSubSum2(int[] a) {
int maxSum = 0;
for(int i = 0; i < a.length; i++) {
//将sumMax放在for循环外面，避免j的变化引起i到j的和sumMax要用for循环重新计算
int sumMax = 0;
for(int j = i; j < a.length; j++) {
sumMax += a[j];
if(sumMax>maxSum) {
maxSum = sumMax;
}
}
}
return maxSum;
}

//递归，分治策略
//2分logn，for循环n，固O(nlogn)
public static int maxSubSum3(int[] a) {
return maxSumRec(a, 0, a.length - 1);
}
public static int maxSumRec(int[] a, int left, int right) {
//递归中的基本情况
if(left == right) {
if(a[left] > 0) return a[left];
else return 0;
}
int center = (left + right) / 2;
//最大子序列在左侧
int maxLeftSum = maxSumRec(a, left, center);
//最大子序列在右侧
int maxRightSum = maxSumRec(a, center+1, right);
//最大子序列在中间（左边靠近中间的最大子序列+右边靠近中间的最大子序列）
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 i = center+1; i<= right; i++) {
rightBorderSum += a[i];
if(rightBorderSum > maxRightBorderSum) maxRightBorderSum = rightBorderSum;
}
//返回最大子序列在左侧，在右侧，在中间求出的值中的最大的
return max3(maxLeftSum, maxRightSum, maxLeftBorderSum + maxRightBorderSum);
}
public static int max3(int a, int b, int c) {
return a > b?(a>c?a:c):(b>c?b:c);
}
//任何a[i]为负时，均不可能作为最大子序列前缀；任何负的子序列不可能是最有子序列的前缀
public static int maxSubSum4 (int [] a) {
int maxSum = 0, thisSum = 0;
for(int j = 0; j < a.length; j++) {
thisSum += a[j];
if(thisSum>maxSum) maxSum = thisSum;
else if (thisSum < 0) thisSum = 0;
}
return maxSum;
}
}