[leetcode]238. Product of Array Except Self
Analysis
new plan~—— [好好养生啊!!]
Given an array nums of n integers where n > 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
因为题目规定不能用出发,所以我们用两个数组,第一个放i左边所有数的乘积,第二个放i右边所有数的乘积。
Implement
方法一(两个数组)
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int len = nums.size();
vector<int> left(len, 1);
vector<int> right(len, 1);
vector<int> res(len, 1);
for(int i=1; i<len; i++)
left[i] = left[i-1]*nums[i-1];
for(int i=len-2; i>=0; i--)
right[i] = right[i+1]*nums[i+1];
for(int i=0; i<len; i++)
res[i] = left[i]*right[i];
return res;
}
};
方法二(一个数组,但道理和方法一一样)
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int len = nums.size();
vector<int> res(len, 1);
for(int i=1; i<len; i++)
res[i] = res[i-1]*nums[i-1];
int tmp = 1;
for(int i=len-1; i>=0; i--){
res[i] *= tmp;
tmp *= nums[i];
}
return res;
}
};