39. Combination Sum

Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

It is guaranteed that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

Example 1:

Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]

Explanation: 2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times. 7 is a candidate, and 7 = 7. These are the only two combinations.

Example 2:

Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3:

Input: candidates = [2], target = 1
Output: []

Example 4:

Input: candidates = [1], target = 1
Output: [[1]]

Example 5:

Input: candidates = [1], target = 2
Output: [[1,1]]

Constraints:

  • 1 <= candidates.length <= 30
  • 1 <= candidatesi <= 200
  • All elements of candidates are distinct.
  • 1 <= target <= 500

analyze

根据题目 return a list of all unique combinations of candidates 可以看出此题属于组合类型问题。

target = 7

  • 2
    • 2
      • 2
        • 2 当前及后续值减支
      • 3 √
      • 6 当前及后续值减支
      • 7
    • 3
      • 3
      • 6
      • 7
    • 6
    • 7
  • 3
    • 3
      • 3
    • 6
    • 7
  • 6
    • 6
  • 7 √

下面给出两个 ac 了的写法, 思路大体相同, 个人推荐拆成两个函数的第二种写法:

写法一:

/**
 * @param {number[]} candidates
 * @param {number} target
 * @return {number[][]}
 */
var combinationSum = function (candidates, target) {
  const result = []

  const DFS = function (sum, arr, start) {
    if (sum === target) {
      result.push(arr.slice())
      return
    }
    if (sum > target) {
      return
    }

    for (let i = start; i < candidates.length; i++) {
      sum += candidates[i]
      arr.push(candidates[i])
      DFS(sum, arr, i)
      arr.pop(candidates[i])
      sum -= candidates[i]
    }
  }

  DFS(0, [], 0, candidates)

  return result
};

写法二:

/**
 * @param {number[]} candidates
 * @param {number} target
 * @return {number[][]}
 */
var combinationSum = function (candidates, target) {
  const result = []
  recurcive(candidates, target, 0, [], result)
  return result
};

var recurcive = function (candidates, target, start, temp, result) {
  if (target < 0) {
    return
  }

  if (target === 0) {
    result.push([...temp])
    return
  }

  for (let i = start; i < candidates.length; i++) {
    temp.push(candidates[i])
    target = target - candidates[i]
    recurcive(candidates, target, i, temp, result)
    target = target + candidates[i]
    temp.pop(candidates[i])
  }
}