LeetCode: Maximum Product Subarray && Maximum Subarray &子序列相关
Maximum Product Subarray
Title:
Find the contiguous subarray within an array (containing at least one number) which has the largest product.
For example, given the array [2,3,-2,4]
,
the contiguous subarray [2,3]
has the largest product = 6
.
对于Product Subarray,要考虑到一种特殊情况,即负数和负数相乘:如果前面得到一个较小的负数,和后面一个较大的负数相乘,得到的反而是一个较大的数,如{2,-3,-7},所以,我们在处理乘法的时候,除了需要维护一个局部最大值,同时还要维护一个局部最小值,由此,可以写出如下的转移方程式:
max_copy[i] = max_local[i]
max_local[i + 1] = Max(Max(max_local[i] * A[i], A[i]), min_local * A[i])
min_local[i + 1] = Min(Min(max_copy[i] * A[i], A[i]), min_local * A[i])
class Solution {
public:
int maxProduct(vector<int>& nums) {
int pmin = nums[];
int pmax = nums[];
int result = nums[];
for (int i = ; i < nums.size(); i++){
int t1= pmax * nums[i];
int t2= pmin * nums[i];
pmax = max(nums[i],max(t1,t2));
pmin = min(nums[i],min(t1,t2));
result = max(result,pmax);
}
return result;
}
};
Maximum Subarray
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
.
http://blog.csdn.net/joylnwang/article/details/6859677
http://blog.csdn.net/linhuanmars/article/details/21314059
class Solution{
public:
int maxSubArray(int A[], int n) {
int maxSum = A[];
int sum = A[];
for (int i = ; i < n; i++){
if (sum < )
sum = ;
sum += A[i];
maxSum = max(sum,maxSum);
}
return maxSum;
}
};
扩展:子序列之和最接近于0
先对数组进行累加,这样得到同样长度的数组,然后,对数组排序,对排序后的数组相邻的元素相减计算绝对值,并比较大小。
class Solution{
public:
vector<int> simple(vector<int> nums,int target){
int min_gap = INT_MAX;
int index_min ;
int index_max;
for (int i = ; i < nums.size(); i++){
int sum = ;
for (int j = i; j < nums.size(); j++){
sum += nums[j];
if (min_gap > abs(sum-target)){
min_gap = abs(sum-target);
index_min = i;
index_max = j;
}
}
}
vector<int> result(nums.begin()+index_min,nums.begin()+index_max+);
return result;
}
vector<int> choose(vector<int> nums, int target){
vector<pair<int,int> > addSums(nums.size());
addSums[] = make_pair(nums[],);
for (int i =; i < nums.size(); i++){
addSums[i] = make_pair(addSums[i-].first + nums[i],i);
}
sort(addSums.begin(),addSums.end());
int min_gap = INT_MAX;
int index = -;
for (int i = ; i < addSums.size(); i++){
int t = abs(addSums[i].first - addSums[i-].first);
if (min_gap > t){
min_gap = t;
index = i;
}
}
int index_min = min(addSums[index].second,addSums[index-].second);
int index_max = max(addSums[index].second,addSums[index-].second);
vector<int> result(nums.begin()+index_min+,nums.begin()+index_max+);
return result;
}
};
这种做法我没有想到如何扩展到任意的t上面
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