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238. Product of Array Except Self
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].
Example:
Input:Output:[1,2,3,4][24,12,8,6]
Note: *Please solve it *without division and in O(n).
Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)
思路分析
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一刷
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/**
* Runtime: 1 ms, faster than 100.00% of Java online submissions for Product of Array Except Self.
*
* Memory Usage: 42.8 MB, less than 48.03% of Java online submissions for Product of Array Except Self.
*
* @author D瓜哥 · https://www.diguage.com
* @since 2020-01-05 20:15
*/
public int[] productExceptSelf(int[] nums) {
int[] result = new int[nums.length];
result[0] = 1;
for (int i = 1; i < nums.length; i++) {
result[i] = nums[i - 1] * result[i - 1];
}
int temp = 1;
for (int i = nums.length - 1; i >= 0; i--) {
result[i] *= temp;
temp *= nums[i];
}
return result;
}
/**
* Runtime: 1 ms, faster than 100.00% of Java online submissions for Product of Array Except Self.
*
* Memory Usage: 42.8 MB, less than 48.03% of Java online submissions for Product of Array Except Self.
*/
public int[] productExceptSelfDivision(int[] nums) {
int product = 1;
int zero = nums.length;
for (int num : nums) {
if (num == 0) {
zero++;
continue;
}
product *= num;
}
int[] result = new int[nums.length];
if (zero >= nums.length + 2) {
return result;
}
for (int i = 0; i < nums.length; i++) {
int num = nums[i];
if (num == 0) {
result[i] = product;
} else {
if (zero == nums.length + 1) {
result[i] = 0;
} else {
result[i] = product / num;
}
}
}
return result;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-09-14 15:53:49
*/
public int[] productExceptSelf(int[] nums) {
// 可以把 pre 和 post 简化掉，直接在 result 上进行计算
int[] pre = new int[nums.length];
pre[0] = nums[0];
for (int i = 1; i < nums.length; i++) {
pre[i] = nums[i] * pre[i - 1];
}
int[] post = new int[nums.length];
post[nums.length - 1] = nums[nums.length - 1];
for (int i = nums.length - 2; i >= 0; i--) {
post[i] = nums[i] * post[i + 1];
}
int[] result = new int[nums.length];
for (int i = 0; i < nums.length; i++) {
int preNum = 1;
if (0 < i) {
preNum = pre[i - 1];
}
int postNum = 1;
if (i < nums.length - 1) {
postNum = post[i + 1];
}
result[i] = preNum * postNum;
}
return result;
}
该题的思路也适用于 724. Find Pivot Index。