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面试题57 - II. 和为s的连续正数序列

题目描述

输入一个正整数 target ，输出所有和为 target 的连续正整数序列（至少含有两个数）。

序列内的数字由小到大排列，不同序列按照首个数字从小到大排列。

示例 1：

输入：target = 9
输出：[[2,3,4],[4,5]]

示例 2：

输入：target = 15
输出：[[1,2,3,4,5],[4,5,6],[7,8]]

限制：

• 1 <= target <= 10^5

解法

双指针：p = 1，q = 2。

Python3

class Solution:
    def findContinuousSequence(self, target: int) -> List[List[int]]:
        res = []
        p, q = 1, 2
        while p < q:
            s = (p + q) * (q - p + 1) >> 1
            if s == target:
                res.append([i for i in range(p, q + 1)])
                p += 1
            elif s < target:
                q += 1
            else:
                p += 1
        return res        

Java

class Solution {
    public int[][] findContinuousSequence(int target) {
        List<int[]> list = new ArrayList<>();
        int p = 1, q = 2;
        while (p < q) {
            int s = (p + q) * (q - p + 1) >> 1;
            if (s == target) {
                int[] t = new int[q - p + 1];
                for (int i = 0; i < t.length; ++i) {
                    t[i] = p + i;
                }
                list.add(t);
                ++p;
            } else if (s < target) {
                ++q;
            } else {
                ++p;
            }
        }
        int[][] res = new int[list.size()][];
        for (int i = 0; i < res.length; ++i) {
            res[i] = list.get(i);
        }
        return res;
    }
}

JavaScript

/**
 * @param {number} target
 * @return {number[][]}
 */
var findContinuousSequence = function(target) {
    let res = []
    let window = []
    let i = 1
    let sum = 0
    while(1) {
        if(sum < target) {
            window.push(i)
            sum += i
            i++
        } else if(sum > target) {
            let a = window.shift()
            if(window.length < 2) break
            sum -= a
        } else {
            res.push([...window])
            window.push(i)
            sum += i
            i++
            if(window.length === 2) break
        }
    }
    return res
};

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