40. Combination Sum II

[Tcdian]

40. Combination Sum II

给定一个数组 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 中的每个数字在每个组合中只能使用一次。

Note

• 所有数字（包括目标数）都是正整数。
• 解集不能包含重复的组合。

Example 1

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

[Tcdian]

Solution

• JavaScript Solution

/**
 * @param {number[]} candidates
 * @param {number} target
 * @return {number[][]}
 */
var combinationSum2 = function(candidates, target) {
    candidates.sort((a, b) => a - b);
    const cache = new Set();
    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) {
                if (!cache.has(part.join(','))) {
                    cache.add(part.join(','));
                    reuslt.push([...part]);
                }
            }
            return;
        }
        for (let i = index; i < candidates.length; i++) {
            part.push(candidates[i]);
            backtracking(i + 1);
            part.pop();
        }
    }
};

• TypeScript Solution

function combinationSum2(candidates: number[], target: number): number[][] {
    candidates.sort((a, b) => a - b);
    const cache: Set<string> = new Set();
    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) {
                if (!cache.has(part.join(','))) {
                    cache.add(part.join(','));
                    reuslt.push([...part]);
                }
            }
            return;
        }
        for (let i = index; i < candidates.length; i++) {
            part.push(candidates[i]);
            backtracking(i + 1);
            part.pop();
        }
    }
};