问题描述
我试图在 Rust 中解决以下 leetcode problem:
impl Solution {
pub fn merge(nums1: &mut Vec<i32>,m: i32,nums2: &mut Vec<i32>,n: i32) {
let mut backwards_idx: usize = (m + n - 1) as usize;
let (m_size,n_size) = (m as usize,n as usize);
let (mut m_idx,mut n_idx) = ((m-1) as usize,(n-1) as usize);
while 0 <= m_idx && 0 <= n_idx {
let (nums1_elem,nums2_elem) = (nums1[m_idx],nums2[n_idx]);
if nums1_elem <= nums2_elem {
nums1[backwards_idx] = nums2_elem;
n_idx = n_idx - 1;
} else {
nums1[backwards_idx] = nums1_elem;
m_idx = m_idx - 1;
}
backwards_idx = backwards_idx - 1;
}
while 0 <= m_idx {
nums1[backwards_idx] = nums1[m_idx];
m_idx = m_idx - 1;
backwards_idx = backwards_idx - 1;
}
while 0 <= n_idx {
nums1[backwards_idx] = nums2[n_idx];
n_idx = n_idx - 1;
backwards_idx = backwards_idx - 1;
}
}
}
然而,这行不通,因为 m_size
和 n_size
从 0 中减去时会溢出。我知道这不是规范的 Rust,但由于问题在 {{1} },我不能改变它。
有 checked method ,但这似乎很难阅读,而不是直接在 while 循环条件中写入它。不知何故,在 while 循环体中设置退出条件似乎不是一个好主意。
解决方法
避免必须检查溢出的一种简单方法是确保您的 usize 类型(只减 1)从 0 开始那些循环。将 idx 类型重构为计数,并在我们想要获得索引时减 1,是一种方法......
impl Solution {
pub fn merge(nums1: &mut Vec<i32>,m: i32,nums2: &mut Vec<i32>,n: i32) {
let mut numbers_left: usize = (m + n) as usize;
let (mut m_left,mut n_left) = (m as usize,n as usize);
while m_left > 0 && n_left > 0 {
let (nums1_elem,nums2_elem) = (nums1[m_left-1],nums2[n_left-1]);
if nums1_elem <= nums2_elem {
nums1[numbers_left-1] = nums2_elem;
n_left = n_left - 1;
} else {
nums1[numbers_left-1] = nums1_elem;
m_left = m_left - 1;
}
numbers_left = numbers_left - 1;
}
while m_left > 0 {
nums1[numbers_left-1] = nums1[m_left-1];
m_left = m_left - 1;
numbers_left = numbers_left - 1;
}
while n_idx > 0 {
nums1[numbers_left-1] = nums2[n_left-1];
n_left = n_left - 1;
numbers_left = numbers_left - 1;
}
}
}
,
一种避免溢出的方法,就是用不同的方式构造循环...... Rust 没有明确的 do-while
语法,但它确实有 loop
,你可以{{1} } from,甚至使用标签突破到任意数量的外部循环。
break