39. Combination Sum

[Tcdian]

39. Combination Sum

给定一个 无重复元素 的数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的数字可以无限制重复被选取。

Note

• 所有数字（包括 target）都是正整数。
• 解集不能包含重复的组合。

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]
]

Constraints

• 1 <= candidates.length <= 30
• 1 <= candidates[i] <= 200
• candidate 中的每个元素都是独一无二的。
• 1 <= target <= 500

[Tcdian]

Solution

• JavaScript Solution

/**
 * @param {number[]} candidates
 * @param {number} target
 * @return {number[][]}
 */
var combinationSum = function(candidates, target) {
    candidates.sort((a, b) => a - b);
    const reuslt = [];
    const part = [];
    backtracking(0);
    return reuslt;

    function backtracking(index) {
        const sum = part.reduce((prev, current) => prev + current, 0);
        if (sum >= target) {
            if (sum === target) {
                reuslt.push([...part]);
            }
            return;
        }
        for (let i = index; i < candidates.length; i++) {
            part.push(candidates[i]);
            backtracking(i);
            part.pop();
        }
    }
};

• TypeScript Solution

function combinationSum(candidates: number[], target: number): number[][] {
    candidates.sort((a, b) => a - b);
    const reuslt: number[][] = [];
    const part: number[] = [];
    backtracking(0);
    return reuslt;

    function backtracking(index: number) {
        const sum = part.reduce((prev, current) => prev + current, 0);
        if (sum >= target) {
            if (sum === target) {
                reuslt.push([...part]);
            }
            return;
        }
        for (let i = index; i < candidates.length; i++) {
            part.push(candidates[i]);
            backtracking(i);
            part.pop();
        }
    }
};