🌕 40. 组合总和 II

难度: 🌕

问题描述

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解法

class Solution {
    List<List<Integer>> res = new LinkedList<>();
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        // 思路：
        // 可重复数组 + 无重复选取
        Arrays.sort(candidates);
        int len = candidates.length;
        LinkedList<Integer> path = new LinkedList<>();
        mySol(candidates, target, len, 0, 0, path);
        return res;
    }

    private void mySol(int[] candidates, int target, int len, int index, int sum, LinkedList<Integer> path) {
        // 递归终止条件
        if(sum > target) {
            return;
        }
        if(sum == target) {
            res.add(new LinkedList<>(path));
            return;
        }
        if(index > len) {
            return;
        }
        for(int i = index; i < len; i ++) {
            if(sum + candidates[i] > target) {
                return;
            }
            // 当前枝若前一个元素已经被选取，则当前枝不存在
            if(i > index && candidates[i] == candidates[i - 1]) {
                continue;
            }
            path.addLast(candidates[i]);
            mySol(candidates, target, len, i + 1, sum + candidates[i], path);
            path.removeLast();
        }
    }
}

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输出