分治法求最大子段和

#include <stdio.h>
int max_sub_sum(int a[],int left, int right){
int center,i,j,sum,left_sum,right_sum,s1,s2,lefts,rights;
if(left == right){
/*二分法递归结束条件*/
if(a[left] > 0)
return a[left];
else
return 0;

}else{
center = (left + right)/2;
left_sum = max_sub_sum(a, left,center); /*求左边最大子段*/
right_sum = max_sub_sum(a, center + 1, right);/*求右边最大子段*/
/*下面求中间交叉部分最大子段*/
s1 = 0;
lefts = 0;
for(i=center; i>=left;i--){
lefts = lefts + a[i];
if(lefts > s1){
s1 = lefts;
}
}
s2 =0;
rights = 0;
for(j = center + 1; j<=right;j++){
rights = rights + a[j];
if(rights > s2){
s2 = rights;
}
}
/*选择最大子段*/
if((s1 + s2 < left_sum) && (right_sum < left_sum))return left_sum;
if(s1 + s2 < right_sum)return right_sum;
return s1+s2;
}

}
int main(){
int a[] = {-2,1,-4,13,-50,6};
int left_partion = -1; /*左边界*/
int right_partion = -1;/*右边界*/
int i =0;
int ret = max_sub_sum(a,1,5);
printf("max_sub_sum is:%d\n",ret);

return 0;
}