[Vssue]0039.组合总和.md

[youngyangyang04]

https://www.programmercarl.com/0039.%E7%BB%84%E5%90%88%E6%80%BB%E5%92%8C.html

[Du1in9]

private void backtracking(int[] nums, int target, int sum, int start) {
    if(sum == target){
        paths.add(new ArrayList<>(path));
        return;
    }
    for(int i = start; i < nums.length && sum + nums[i] <= target; i++){
        sum += nums[i];
        path.add(nums[i]);
        backtracking(nums, target, sum, i);
        sum -= nums[i];
        path.remove(path.size() - 1);
    }
}

// 深度1: backtracking([2,3,6,7], 7, 0, 0), 在 [2,3,6,7] 中取一个, path = []
i = 0：path = [2]，递归 backtracking([2,3,6,7], 7, 0, 0)，回溯，path = [] ...
i = 3：path = [7]，递归 backtracking([2,3,6,7], 7, 0, 3)，回溯，path = []   
// 深度2: 例: backtracking([2,3,6,7], 7, 0, 0), 在 [2,3,6,7] 中取一个, path = [2]
i = 0：path = [2,2]，递归 backtracking([2,3,6,7], 7, 0, 0)，回溯，path = [2]
i = 1：path = [2,3]，递归 backtracking([2,3,6,7], 7, 0, 1)，回溯，path = [2]
i = 2：不满足 2 + 6 <= 7, 停止遍历（剪枝）
// 深度3: 例: backtracking([2,3,6,7], 7, 0, 0), 在 [2,3,6,7] 中取一个, path = [2,2]
i = 0：path = [2,2,2]，递归 backtracking([2,3,6,7], 7, 0, 0)，回溯，path = [2,2]
i = 1：path = [2,2,3]，递归 backtracking([2,3,6,7], 7, 0, 1，回溯，path = [2,2]
i = 2：不满足 4 + 6 <= 7, 停止遍历（剪枝）
// 深度n: 以此类推