[Vssue]0491.递增子序列.md

[youngyangyang04]

https://www.programmercarl.com/0491.%E9%80%92%E5%A2%9E%E5%AD%90%E5%BA%8F%E5%88%97.html

[muyiGin]

我爱你Carl，谢谢你无偿分享好知识

[Du1in9]

private void backtracking(int[] nums, int start) {
    if(path.size() >= 2){
        paths.add(new ArrayList<>(path));
    }
    Set<Integer> set = new HashSet<>();
    for(int i = start; i < nums.length; i++){
        if(!path.isEmpty() && nums[i] < path.get(path.size() - 1)){                 
            continue;               // 保证递增
        }
        if(set.contains(nums[i])) continue; // 去重
        set.add(nums[i]);
        path.add(nums[i]);
        backtracking(nums, i + 1);
        path.remove(path.size() - 1);
    }
}

// 深度1: backtracking([4,7,6,7], 0), path = []
i = 0：path = [4]，递归 backtracking([4,6,7,7], 1)。回溯，path = []
i = 1：path = [7]，递归 backtracking([4,6,7,7], 2)。回溯，path = []
i = 2：path = [6]，递归 backtracking([4,6,7,7], 3)。回溯，path = []
i = 3：满足 set.contains(7)，继续遍历（去重）
// 深度2: 例: backtracking([4,7,6,7], 1), path = [4]
i = 1：path = [4,7]，递归 backtracking([4,6,7,7], 2)，加入 paths。回溯，path = [4]
i = 2：path = [4,6]，递归 backtracking([4,6,7,7], 3)，加入 paths。回溯，path = [4]
i = 3：满足 set.contains(7)，继续遍历（去重）
// 深度3: 例: backtracking([4,7,6,7], 2), path = [4,7]
i = 2：满足 6 < 7，继续遍历（保证递增）
i = 3：path = [4,7,7]，递归 backtracking([4,6,7,7], 4)，加入 paths。回溯，path = [4,7]