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].
题意:给定一个数组,在每个位置生成其他位置相乘得到的值。
思路:定义一个值记录0的个数,一个值记录除0外元素的乘积。
时间复杂度O(n),空间复杂度O(1)。
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int temp = 1, zeroCount = 0;
for (auto ele : nums) {
if (ele) temp *= ele;
else zeroCount++;
}
if(zeroCount == 0)
for (int i = 0; i < nums.size(); ++i)
nums[i] = temp / nums[i];
else
for (int i = 0; i < nums.size(); ++i) {
if (nums[i]) nums[i] = 0;
else if (zeroCount == 1) nums[i] = temp;
else nums[i] = 0;
}
return nums;
}
};