题目:

iven 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:

这道题只要将nums中乘积求出来,然后将乘积除以nums中每一项(nums[i])即可。但要注意0的情况。

/**
* @param {number[]} nums
* @return {number[]}
*/
var productExceptSelf = function(nums) {
var pro=1,res=[],flag=0;
for(var i=0,len=nums.length;i<len;i++){
if(nums[i]==0){
flag++;
}else{
pro*=nums[i];
}
} for(var i=0,len=nums.length;i<len;i++){
if(nums[i]!=0&&flag){
res[i]=0;
}else if(nums[i]!=0&&flag==0){
res[i]=pro/nums[i]
}else if(nums[i]==0&&flag==1){
res[i]=pro;
}else if(nums[i]==0&&flag>1){
res[i]=0;
}
}
return res;
};

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