题目描述
暴力法
class Solution {
public int maxSubArray(int[] a) {
int n = a.length;
if(n==1){
return a[0];
}
int max=Integer.MIN_VALUE,esti=0;
for(int i=0;i<n;i++){
//以a[i]开头的子段求和，取最大值
int sub= 0;
for(int j=i;j<n;j++){
//a[i]~a[j]子段
sub+=a[j];
if(sub>max){
max = sub;
}
}
}
return max;
}
}
分治法
class Solution {
public int maxSubArray(int[] a) {
return MaxSubsum(a,0,a.length-1);
}
public int MaxSubsum(int []a,int left,int right){
int sum;
if(left==right)
sum=a[left];
else{
//位于左边的一半|或者位于右边的一半
int center = (left+right)/2;
int leftsum = MaxSubsum(a,left,center);
int rightsum = MaxSubsum(a,center+1,right);
//位于中间部分
//先求左边的最大值
int s1=0,lefts=Integer.MIN_VALUE;
for(int i=center;i>=left;i--){
s1+=a[i];
if(lefts<s1){
lefts = s1;
}
}
//右边最大值
int s2=0,rights=Integer.MIN_VALUE;
for(int i=center+1;i<=right;i++){
s2+=a[i];
if(rights<s2){
rights = s2;
}
}
//求三者中的最大值
sum = rights+lefts;
if(sum<leftsum) sum=leftsum;
if(sum<rightsum) sum = rightsum;
}
return sum;
}
}
动态规划
class Solution {
public int maxSubArray(int[] a) {
int n = a.length;
int []b = new int[n];
b[0]=a[0];
int max=b[0];
for(int i=1; i<n; i++)
{
if(b[i-1]>0)
b[i]=b[i-1]+a[i];
else
b[i]=a[i];
if(b[i]>max)
max=b[i];
}
return max;
}
思路解析请参考我的博客：算法学习（三）动态规划（重要）