最大子序列的3种不同时间复杂度的算法

第一种是最挫的O(N^2)的算法，就是直接暴力二重循环解之。

public class MaxSubSum1 {
public static int maxSubSum2(int[] a){
int maxSum = 0;
for(int i = 0;i < a.length;i ++){
int thisSum = 0;
for(int j = i;j < a.length;j ++){
thisSum += a[j];
if(thisSum > maxSum)
maxSum = thisSum;
}
}
return maxSum;
}
public static void main(String[] args){
int[] a = {-2,11,-4,13,-5,-2};
int maxSum = maxSubSum1(a);
System.out.println(maxSum);
}
}

第二种的时间复杂度是O(NlogN),用到递归的方法，分治的思想，如果没有后面的的O(N)的算法，这个算法便体现了递归的威力。

public class MaxSubSum2 {
public static int maxSumRec(int[] a,int left,int right){
/*当这个数组只有一个元素的时候*/
if(left == right)
if(a[left] > 0)
return a[left];
else
return 0;
/*递归*/
int center = (left + right) >> 1;
int maxLeftSum = maxSumRec(a,left,center);
int maxRightSum = maxSumRec(a, center+1, right);
/*求左段的最大子序列*/
int maxLeftBorderSum = 0,leftBorderSum = 0;
for(int i = center;i >= left;i --){
leftBorderSum += a[i];
if(leftBorderSum > maxLeftBorderSum)
maxLeftBorderSum = leftBorderSum;
}
/*求右段的最大子序列*/
int maxRightBorderSum = 0,rightBorderSum = 0;
for(int i = center + 1;i <= right;i ++){
rightBorderSum += a[i];
if(rightBorderSum > maxRightBorderSum)
maxRightBorderSum = rightBorderSum;
}
/*返回左段，右段，中间段三者的最大值*/
return max3(maxLeftSum,maxRightSum,maxLeftBorderSum + maxRightBorderSum);
}
/*max3方法*/
private static int max3(int maxLeftSum, int maxRightSum, int i) {
if(maxLeftSum >= maxRightSum && maxLeftSum >= i)
return maxLeftSum;
if(maxRightSum >= maxLeftSum && maxRightSum >= i)
return maxRightSum;
if(i >= maxRightSum && i >= maxLeftSum)
return i;
else
return i;
}

public static int maxSubSum2(int[] a){
return maxSumRec(a, 0, a.length - 1);
}

public static void main(String[] args){
int[] a =  {4,-3,5,-2,-1,2,6,-2};
System.out.println(maxSubSum2(a));
}
}

第三种是最聪明也是最简单的方法了，时间复杂度为O(N)，代码简单易懂，用thisSum来记录每一次累加的结果，如果大于maxSum，就让maxSum = thisSum，如果thisSum小于0,便让thisSum重新清零，如此从头遍历到尾即可。

public class MaxSubSum3 {
public static int maxSubSum3(int[] a){
int maxSum = 0,thisSum = 0;
for(int i = 0;i < a.length;i ++){
thisSum += a[i];

if(thisSum > maxSum)
maxSum = thisSum;
if(thisSum < 0)
thisSum = 0;
}
return maxSum;
}

public static void main(String[] args){
int[] a = {4,-3,5,-2,-1,2,6,-2};
System.out.println(maxSubSum3(a));
}
}