題目來(lái)源
Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
Solve it without division and in O(n).
For example, given [1,2,3,4], return [24,12,8,6].
Follow up:Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)
這道題看到的第一想法就是先把所有相乘棠赛，然后再除以每個(gè)nums[i]戏羽，然后還得處理一些特殊情況钱骂。
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
long long sum = 1;
vector<int> res;
int count0 = 0;
for (auto num : nums) {
if (num == 0) {
count0++;
continue;
}
sum *= num;
}
if (count0 > 1)
return vector<int>(nums.size(), 0);
for (auto num : nums) {
if (count0 == 0)
res.push_back(sum / num);
else {
if (num == 0)
res.push_back(sum);
else
res.push_back(0);
}
}
return res;
}
};
感覺(jué)代碼寫的比較亂…然后我還是想的太多，直接乘不好嗎對(duì)吧…代碼如下：
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int n = nums.size();
vector<int> res(n, 1);
for (int i=1; i<n; i++) {
res[i] = res[i-1] * nums[i-1];
}
int a = 1;
for (int i=n-1; i>0; i--) {
a *= nums[i];
res[i-1] *= a;
}
return res;
}
};
