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Commit 8dff829

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Solve #238
1 parent e4ab1e0 commit 8dff829

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‎src/lib.rs

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@@ -200,3 +200,5 @@ mod n0229_majority_element_ii;
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mod n0230_kth_smallest_element_in_a_bst;
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mod n0231_power_of_two;
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mod n0232_implement_queue_using_stacks;
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mod n0233_number_of_digit_one;
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mod n0238_product_of_array_except_self;

‎src/n0233_number_of_digit_one.rs

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/**
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* [233] Number of Digit One
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*
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* Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.
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*
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* Example:
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*
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*
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* Input: 13
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* Output: 6
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* Explanation: Digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.
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*
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*
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*/
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pub struct Solution {}
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// submission codes start here
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impl Solution {
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pub fn count_digit_one(n: i32) -> i32 {
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0
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}
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}
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// submission codes end
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#[cfg(test)]
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mod tests {
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use super::*;
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#[test]
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fn test_233() {}
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}
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/**
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* [238] Product of Array Except Self
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*
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* 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].
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*
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* Example:
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*
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*
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* Input: [1,2,3,4]
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* Output: [24,12,8,6]
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*
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*
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* Note: Please solve it without division and in O(n).
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*
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* Follow up:<br />
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* Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)
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*
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*/
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pub struct Solution {}
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// submission codes start here
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/*
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x 2 3 4 = 24
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1 x 3 4 = 12
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1 2 x 4 = 8
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1 2 3 x = 6
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*/
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impl Solution {
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pub fn product_except_self(nums: Vec<i32>) -> Vec<i32> {
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if nums.len() < 2 { return vec![] }
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let mut res = vec![1; nums.len()];
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let mut n = 1;
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for i in (0..nums.len()-1).rev() {
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n *= nums[i+1];
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res[i] = n;
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}
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n = 1;
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for i in 1..nums.len() {
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n *= nums[i-1];
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res[i] *= n;
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}
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res
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}
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}
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// submission codes end
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#[cfg(test)]
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mod tests {
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use super::*;
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#[test]
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fn test_238() {
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assert_eq!(Solution::product_except_self(vec![1,2,3,4]), vec![24,12,8,6]);
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}
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}

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