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reverse_pairs.rs
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// 翻转对
// https://leetcode.cn/problems/reverse-pairs/
// INLINE ../../images/array/reverse_pairs.jpeg
// 解题思路:基于并归排序作计数。
pub struct Solution;
impl Solution {
// 合并两个有序数组
fn merge(nums: &mut Vec<i32>, left: usize, middle: usize, right: usize) {
let mut i = left;
let mut j = middle + 1;
let mut k = left;
let mut temp = vec![];
while k <= right {
if i > middle {
temp.push(nums[j]);
j += 1;
k += 1;
} else if j > right {
temp.push(nums[i]);
i += 1;
k += 1;
} else if nums[i] < nums[j] {
temp.push(nums[i]);
i += 1;
k += 1;
} else {
temp.push(nums[j]);
j += 1;
k += 1;
}
}
// 将排序后的数组复制回原数组
for i in 0..=(right - left) {
nums[left + i] = temp[i];
}
}
// 计算翻转对的数量
fn merge_sort_recursion_count(nums: &mut Vec<i32>, left: usize, right: usize) -> usize {
// 递归终止条件
if left >= right {
return 0;
}
let middle = (left + right) >> 1;
// 先统计左、右子数组的重要翻转对数量
let mut count = Self::merge_sort_recursion_count(nums, left, middle)
+ Self::merge_sort_recursion_count(nums, middle + 1, right);
// 再统计左、右子数组之间的重要翻转对数量
let mut i = left;
let mut j = middle + 1;
while i <= middle && j <= right {
// 防止溢出
if nums[i] as i64 > 2 * nums[j] as i64 {
count += middle - i + 1;
j += 1;
} else {
i += 1;
}
}
// 合并两个有序数组
Self::merge(nums, left, middle, right);
count
}
pub fn reverse_pairs(mut nums: Vec<i32>) -> i32 {
let len = nums.len();
if len <= 1 {
return 0;
}
Self::merge_sort_recursion_count(&mut nums, 0, len - 1) as i32
}
}