leetcode || 53、Maximum Subarray
problem:
Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
For example, given the array [−2,1,−3,4,−1,2,1,−5,4]
,
the contiguous subarray [4,−1,2,1]
has the largest sum = 6
.
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
thinking:
(1)这道题解法特别多:
方法1:将每个数和后一个数字相加,得到一个正负分布的序列,正数项对最大和子序列实用,作比較就可以
方法2:暴力匹配,两层循环,调用max()函数。时间复杂度O(n*n),也能够算出结果,可是提交超时
方法3:採用DP。时间复杂度O(n)
方法4:分治法。时间复杂度为nlog(n)
(2)本人实现了方法3 和方法4
code:
DP 法:
class Solution {
public:
int maxSubArray(int A[], int n) {
int sum=A[0];
int maxsum=A[0];
for(int i=1;i<n;i++)
{
if(sum<0) //DP核心
sum=0;
sum+=A[i];
maxsum=max(sum,maxsum);
}
return maxsum;
} };
分治法:
class Solution {
public:
int maxSubArray(int A[], int n) {
int ret=maxsub(A,0,n-1);
return ret; }
protected:
int maxsub(int A[], int start, int end)
{
int max_left=INT_MIN,max_mid=INT_MIN,max_right=INT_MIN;
if(start==end)
return A[start];
if(start+1==end)
{
int a=max(A[start],A[end]);
return a>(A[start]+A[end])?a:(A[start]+A[end]);
}
int mid=(start+end)/2;
int tmp_sum=A[mid];
max_mid=tmp_sum; int i=mid-1;
int j=mid+1;
while(i>=start) //难点在于当连续最大和子序列分布在mid的一側或两側时,怎么处理
{ tmp_sum+=A[i];
i--;
max_mid=max(max_mid,tmp_sum); }
if(max_mid>A[mid]) //推断是处于两側,还是处于一側
tmp_sum=max_mid;
else
tmp_sum=A[mid];
while(j<=end)
{ tmp_sum+=A[j];
j++;
max_mid=max(max_mid,tmp_sum); } max_left=max(max_left,maxsub(A,start,mid-1));//二分轮廓
max_right=max(max_right,maxsub(A,mid+1,end));
int tmp_max = max(max_left,max_right);
return max_mid>tmp_max?max_mid:tmp_max;
} };
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