152 乘積最大子數組
來源 LeetCode :乘積最大子數組
非動態規劃
一開始我並沒有用DP
而是分情況討論的,即通過0將數據分爲幾段,因爲含有0的必定會使結果爲0
然後就是考慮負數
在所分的每段裏計算乘積,如果負數個數爲奇數,將最後出現的那段除掉,然後輸出最大值再比較
代碼非常長
但是可以跑過
長的我自己都不想看,沒臉看,看不明白啊,孃的
class Solution {
public:
int maxProduct(vector<int>& nums) {
int len = nums.size();
int rst = 0;
if(len == 0)
rst = 0;
else if(len == 1)
rst = nums[0];
else
{
int tag = true;
int i = 0;int j = 0;
for(; j<len; ++j)
{
if(nums[j] == 0)
{
rst = i==j ? rst : max(rst, paraProduct(nums, i, j));
i = j;
++i;
}
}
rst = i == j ? rst : max(rst, paraProduct(nums, i, j));
}
return rst;
}
int paraProduct(vector<int>& nums, int low, int high)
{
if(low == high-1){return nums[low];}
else
{
int neg_head = 1;
int neg_tail = 1;
int count = 0;
int mulp = 1;
int head_mulp = nums[low];
int tail_mulp = nums[high-1];
for(int i=low; i<high; ++i)
{
if(nums[i]<0)
{
tail_mulp = mulp;
if(count < 1)
{
neg_head = nums[i];
++count;
head_mulp = mulp;
}
neg_tail = nums[i];
}
mulp *= nums[i];
}
return max(mulp, max(max(head_mulp, mulp/(neg_head*head_mulp)),max(tail_mulp, mulp/(tail_mulp*neg_tail))));
}
}
};
```
## DP