求给定数组中 N 个数相加之和为 sum 所有可能集合
更多描述 求给定数组中 N 个数相加之和为 sum 所有可能集合,请补充以下代码
function fn(arr, n, sum) {}
Issue 欢迎在 Gtihub Issue 中回答此问题: Issue 692
Author 回答者: shfshanyue
TODO
Author 回答者: heretic-G
function fun(arr, n, sum) {
let result = [];
if (arr.length < n) return -1;
arr.sort((prev, next) => {
return prev - next;
});
function getSum(arr, n, currSum, index, incArr = []) {
for (let i = index; i < arr.length; i++) {
let temp = currSum + arr[i];
if (temp > sum) break;
if (n > 1) {
getSum(arr, n - 1, temp, i + 1, [arr[i], ...incArr]);
}
if (n === 1 && temp === sum) {
result.push([arr[i], ...incArr]);
}
}
}
getSum(arr, n, 0, 0);
return result;
}
Author 回答者: haotie1990
function findSumNumbers(array, n, sum) {
// 枚举所有n个数的组合,判断组合的和等于sum
let result = [];
const generateAll = function (index, collection, arr) {
if (collection.length === n) {
const s = collection.reduce((acc, c) => (acc += c), 0);
if (s === sum) {
result.push(collection);
}
return;
}
for (let i = 0; i < arr.length; i++) {
generateAll(index + 1, collection.concat(arr[i]), arr.slice(i + 1));
}
};
generateAll(0, [], array.slice(0));
return result;
}
findSumNumbers([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 2, 10);
findSumNumbers([1, 0, -1, 0, -2, 2], 4, 0);
Author 回答者: shen076
https://leetcode.cn/problems/combination-sum-ii/
Author 回答者: JamiLanister
function fun(arr, n, sum) {
let ans = []
let combination = []
function handle(start, array) {
ans.push([...array]);
for (let i=start; i<arr.length; i++) {
array.push(arr[i])
handle(i+1, array)
array.pop()
}
}
handle(0, [])
function add(list) {
return list.reduce((prev,cur) => prev+cur, 0)
}
for (let key of ans) {
if (key.length === n && add(key) === sum) {
combination.push([...key])
}
}
return combination
}
console.log(fun([2,3,6, -1,7,-2,9], 2, 5))
Author 回答者: JLUssh
利用到两个元素相加为target,然后在此基础上使用变成三个元素、四个元素…