每周算法:找出指定求和的组合

leetcode第39题,难度为medium,考察回溯算法(backtracking)的使用。

Description

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

  • All numbers (including target) will be positive integers.
  • The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

来源:leetcode 39 combination sum

Solution

Approach 1 brute force

使用暴力解法求解。

bool sumEqual(vector<int>& nums, int target) {
    for (int n : nums) {
	target -= n;
	if (target < 0) {
	    return false;
	}
    }
    return target == 0;
}

vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
    set<vector<int>> ans;
    sort(candidates.begin(), candidates.end());

    // 将所有可能的组合都列举出来
    bool f = target%candidates.front() == 0;
    int lenMax = target/candidates.front() + (f ? 0 : 1); // 可能的最大长度

    list<vector<int>> bfs;
    bfs.push_back(vector<int>());
    while(!bfs.empty()) {
	vector<int> temp = bfs.front();
	bfs.pop_front();
	if (temp.size() >= size_t(lenMax)) {
	    if (sumEqual(temp, target)) {
		ans.insert(temp);
	    }
	    continue;
	}
	for (int n : candidates) {
	    if (!temp.empty() && temp.back() > n) {
		if (sumEqual(temp, target)) {
		    ans.insert(temp);
		}
		continue;
	    }
	    temp.push_back(n);
	    bfs.push_back(temp);
	    temp.pop_back();
	}
    }
    return vector<vector<int>>(ans.begin(), ans.end());
}

Approach 2 traceback

这道题使用回溯算法(backtraking)求解比较合适。

static void helper(const std::vector<int>& cand,
		   int target,
		   std::vector<std::vector<int>>& ans,
		   std::vector<int>& combi,
		   int sum,
		   int pos) {
    if (sum > target) {
	return;
    }
    else if (sum == target) {
	ans.push_back(combi);
	return;
    }

    for (size_t i=pos; i<cand.size(); ++i) {
	combi.push_back(cand[i]);
	helper(cand, target, ans, combi, sum+cand[i], i);
	combi.pop_back();
    }
}

std::vector<std::vector<int>> combinationSum(std::vector<int>& cand, int target) {
    std::sort(cand.begin(), cand.end());
    std::vector<std::vector<int>> ans;
    std::vector<int> combi;
    helper(cand, target, ans, combi, 0, 0);
    return ans;
}

N-Queen problem

回溯算法的最典型的应用是八皇后问题,下面就是八皇后问题的解法代码。使用回溯算法求解N皇后问题的时间复杂度为 O(N!)

bool solveNQUtil(int board[N][N], int col) {
    /* base case: If all queens are placed then return true */
    if (col >= N)
	return true;

    /* Consider this column and try placing this queen in all rows one by one */
    for (int i = 0; i < N; i++) {
	/* Check if queen can be placed on board[i][col] */
	if ( isSafe(board, i, col) ) {
	    /* Place this queen in board[i][col] */
	    board[i][col] = 1;

	    /* recur to place rest of the queens */
	    if ( solveNQUtil(board, col + 1) == true )
		return true;

	    /* If placing queen in board[i][col] doesn't lead to a solution then remove queen from board[i][col] */
	    board[i][col] = 0; // BACKTRACK
	}
    }
}

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