Question

例如：给定一个数组：

var a = [5,4,3,2,1,4,5,6,7,1];

预期总和为：3 + 2 + 6 + 7 = 18

整个代码中应该只有一个for循环。

我失败的方法：

var a = [1,1,1,2,2,2,3,4,5,6,7,7,8,2]
var expected = 3+4+5+6+8

const sum1 = (a) => {
    var s = 0;
    var ch = {};
    for(let i = 0; i<a.length; i++) {
        if(s <0) {
            s = 0;
        }
        if(ch[a[i]]) {
            s -= a[i]
        } else {
            ch[a[i]] = a[i];
            s += a[i];
        }
    }

    return s
}

console.log(sum1(a))

Answer 1

在每次迭代中，将要迭代的数字加到总和上。使用Set来跟踪到目前为止已出现的数字。如果之前一次看到过该数字，则将其从总和中减去（这样就好像该数字从未被添加到总和中一样）。如果被发现两次或更多次，请忽略它，因为已经从总和中减去了它：

const sumUniques = (input) => {
  let sum = 0;
  // duplicates get put into seenOnce
  // and subtracted from the sum
  const seenOnce = new Set();
  // duplicates in seenTwice get completely ignored;
  // they've already been subtracted from the sum
  const seenTwice = new Set();
  
  input.forEach((num) => {
    if (seenTwice.has(num)) {
      return;
    }
    if (seenOnce.has(num)) {
      sum -= num;
      seenTwice.add(num);
    } else {
      sum += num;
      seenOnce.add(num);
    }
  });
  return sum;
}


console.log(sumUniques([5,4,3,2,1,4,5,6,7,1]))
console.log(sumUniques([1,1,1,2,2,2,3,4,5,6,7,7,8,2]))

Answer 2

这是一个使用映射的解决方案，其中键是元素，值是遇到元素的次数的计数。

const sum = array => {
    const map = new Map();
    return array.reduce((sum, e) => {
      const seenCount = (map.get(e) || 0) + 1;
      map.set(e, seenCount);
      if (seenCount === 1) {
        sum += e;
      } else if (seenCount === 2) {
        sum -= e;
      }
      return sum;
    }, 0);
}

Answer 3

Java解决方案

public static void main(String[] args) {

    int[] arr = {5,4,3,2,1,4,5,6,7,1,5};
    int sum = 0;

    Set<Integer> list = new HashSet<>();
    Set<Integer> doneList = new HashSet<>();

    for (int i = 0; i < arr.length; i++) {

        if(doneList.contains(arr[i])) continue;

        if (!list.contains(arr[i])) {
            list.add(arr[i]);
            sum += arr[i];
        } else{
            sum -= arr[i];
            doneList.add(arr[i]);
        }
    }

    System.out.println(sum);

}

使用单个循环仅添加数组的不同元素

3 个答案: