238 Product of Array Except Self

Author: int32bit

README.md

@@ -132,6 +132,7 @@
 + [235 Lowest Common Ancestor of a Binary Search Tree(LCA, 最低公共祖先,二叉搜索树)](algorithms/LowestCommonAncestorofaBinarySearchTree)
 + [236 Lowest Common Ancestor of a Binary Tree(LCA,最低公共祖先，二叉树](algorithms/LowestCommonAncestorofaBinaryTree)
 + [237 Delete Node in a Linked List(O(1)删除单链表节点)](algorithms/DeleteNodeinaLinkedList)
++ [238 Product of Array Except Self(数组操作)](algorithms/ProductofArrayExceptSelf)
 
 ## Database
 

algorithms/ProductofArrayExceptSelf/README.md

@@ -0,0 +1,71 @@
+## Product of Array Except Self 
+
+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.)
+
+## Solution
+
+简单方法是使用两个数组a,b，其中一个存储`nums[0..i - 1]`, 另一个数组存`nums[i + 1..n-1]`, 于是结果为`result[i] = a[i] * b[n - i - 1]`，即
+a存储i前面的积，b存储i后面元素的积
+
+```cpp
+vector<int> productExceptSelf(vector<int>& nums) {
+    if (nums.empty()) // 空
+	    return vector<int>();
+    if (nums.size() == 1) { //只有一个元素，返回{0}
+	    return vector<int>({0});
+    }
+    auto n = nums.size();
+    vector<int> a(n, 1), b(n, 1);
+    for (auto i = 1; i < n; ++i) {
+	    a[i] = a[i - 1] * nums[i - 1];
+	    b[i] = b[i - 1] * nums[n - i];
+    }
+    for (auto i = 0; i < n; ++i) {
+	    a[i] *= b[n - i - 1];
+    }
+    return a;
+}
+```
+
+以上方法O(n)时间，除了结果空间，还需要额外空间O(n)
+
+题目要求O(1)空间。
+
+我们发现a数组得到i前面的积后，只需要累乘后面的积即可。
+
+* `a[n- 1]`不需要乘
+* `a[n - 2] = a[n - 2] * nums[n - 1]`
+* `a[n - 3] = a[n - 3] * nums[n - 1] * nums[n - 2]`
+* ...
+* `a[0] = a[0] * nums[n - 1] * nums[n - 2] * ... * nums[1]`
+
+使用一个变量product，表示从后往前的累乘，则`a[i] = a[i] * product`
+
+```cpp
+vector<int> productExceptSelf(vector<int> &nums) {
+    if (nums.empty()) // 空
+	    return vector<int>();
+    if (nums.size() == 1) { //只有一个元素，返回{0}
+	    return vector<int>({0});
+    }
+    int n = nums.size();
+    vector<int> result(n, 1);
+    for (int i = 1; i < n; ++i)
+	    result[i] = result[i - 1] * nums[i - 1];
+    int product = 1;
+    for (int i = n - 2; i >= 0; --i) {
+	    product *= nums[i + 1];
+	    result[i] *= product;
+    }
+    return result;
+
+}
+```
+此时空间为O(1)

algorithms/ProductofArrayExceptSelf/solve.cpp

@@ -0,0 +1,57 @@
+#include <vector>
+#include <iostream>
+#include <cstdio>
+#include <algorithm>
+using namespace std;
+static inline void print(const vector<int> &a) {
+	for_each(begin(a), end(a), [](int i){cout << i << ' ';});
+	cout << endl;
+}
+class Solution {
+public:
+    /*
+    vector<int> productExceptSelf(vector<int>& nums) {
+	    if (nums.empty()) // 空
+		    return vector<int>();
+	    if (nums.size() == 1) { //只有一个元素，返回{0}
+		    return vector<int>({0});
+	    }
+	    auto n = nums.size();
+	    vector<int> a(n, 1), b(n, 1);
+	    for (auto i = 1; i < n; ++i) {
+		    a[i] = a[i - 1] * nums[i - 1];
+		    b[i] = b[i - 1] * nums[n - i];
+	    }
+	    for (auto i = 0; i < n; ++i) {
+		    a[i] *= b[n - i - 1];
+	    }
+	    return a;
+    }
+    */
+    vector<int> productExceptSelf(vector<int> &nums) {
+	    if (nums.empty()) // 空
+		    return vector<int>();
+	    if (nums.size() == 1) { //只有一个元素，返回{0}
+		    return vector<int>({0});
+	    }
+	    int n = nums.size();
+	    vector<int> result(n, 1);
+	    for (int i = 1; i < n; ++i)
+		    result[i] = result[i - 1] * nums[i - 1];
+	    int product = 1;
+	    for (int i = n - 2; i >= 0; --i) {
+		    product *= nums[i + 1];
+		    result[i] *= product;
+	    }
+	    return result;
+
+    }
+};
+int main(int argc, char **argv)
+{
+	Solution solution;
+	vector<int> v = {1,2,3,4};
+	auto result = solution.productExceptSelf(v);
+	print(result);
+	return 0;
+}