39. Combination Sum

Problem:

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:

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]
]

Solution:

DFS + Backtracking.

To prevent duplications, only loop the right side of the candidates.

/**
 * @param {number[]} candidates
 * @param {number} target
 * @return {number[][]}
 */
let combinationSum = function (candidates, target) {
  return dfs(candidates, target, [], [], 0);
};

function dfs(candidates, target, result, path, start) {
  for (let i = start; i < candidates.length; i++) {
    const cand = candidates[i];

    if (cand > target) {
      continue;
    }

    path.push(cand);
    if (cand === target) {
      result.push(path.slice());
    } else {
      dfs(candidates, target - cand, result, path, i);
    }
    path.pop(cand);
  }

  return result;
}

: .。. o(≧▽≦)o .。.:☆☆: .。. o(≧▽≦)o .。.:☆☆: .。. o(≧▽≦)o .。.:



: .。. o(≧▽≦)o .。.:☆☆: .。. o(≧▽≦)o .。.: