(leetcode) 238. Product of Array Except Self My Submissions QuestionEditorial Solution
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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.)
思路:如果这个数组中有大于等于2个0的时候,肯定都是0,如果这个数组中只有一个0,那么除这个为0的位置外,别的位置都为0,对于没有0的情况,先把所以数存起来,当前位置的值,等于乗起来的值除以当前的值。
代码如下(已通过leetcode)
public class Solution {
public int[] productExceptSelf(int[] nums) {
int count0=0;
for(int i=0;i<nums.length;i++)
if(nums[i]==0) count0++;
if(count0>=2) {
for(int i=0;i<nums.length;i++)
nums[i]=0;
} else {
if(count0==1) {
int tempmul=1;
for(int i=0;i<nums.length;i++)
if(nums[i]!=0) tempmul=tempmul*nums[i];
for(int i=0;i<nums.length;i++)
if(nums[i]==0) nums[i]=tempmul;
else nums[i]=0;
} else {
int tempmul=1;
for(int i=0;i<nums.length;i++)
tempmul=tempmul*nums[i];
for(int i=0;i<nums.length;i++)
nums[i]=tempmul/nums[i];
}
}
return nums;
}
}
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