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/*
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* File: permutations_ii.rs
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* Created Time: 2023-07-15
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* Author: codingonion (coderonion@gmail.com)
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*/
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use std::collections::HashSet;
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/* 回溯算法:全排列 II */
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fn backtrack(mut state: Vec<i32>, choices: &[i32], selected: &mut [bool], res: &mut Vec<Vec<i32>>) {
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// 当状态长度等于元素数量时,记录解
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if state.len() == choices.len() {
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res.push(state);
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return;
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}
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// 遍历所有选择
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let mut duplicated = HashSet::<i32>::new();
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for i in 0..choices.len() {
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let choice = choices[i];
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// 剪枝:不允许重复选择元素 且 不允许重复选择相等元素
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if !selected[i] && !duplicated.contains(&choice) {
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// 尝试:做出选择,更新状态
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duplicated.insert(choice); // 记录选择过的元素值
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selected[i] = true;
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state.push(choice);
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// 进行下一轮选择
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backtrack(state.clone(), choices, selected, res);
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// 回退:撤销选择,恢复到之前的状态
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selected[i] = false;
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state.remove(state.len() - 1);
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}
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}
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}
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/* 全排列 II */
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fn permutations_ii(nums: &mut [i32]) -> Vec<Vec<i32>> {
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let mut res = Vec::new();
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backtrack(Vec::new(), nums, &mut vec![false; nums.len()], &mut res);
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res
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}
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/* Driver Code */
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pub fn main() {
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let mut nums = [ 1, 2, 2 ];
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let res = permutations_ii(&mut nums);
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println!("输入数组 nums = {:?}", &nums);
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println!("所有排列 res = {:?}", &res);
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}
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