面試題57-題目二：和為S的連續正數序列
/*
* 面試題57-題目二：和為S的連續正數序列
* 題目：輸出所有和為S的連續正數序列。序列內按照從小至大的順序，序列間按照開始數字從小到大的順序
* 思路：定義兩個指標，分別遞增，尋找和為s的序列
* 用兩個數字small和big分別表示序列的最大值和最小值，首先將small初始化為1，big初始化為2.
* 如果從samll到big的和大於s，我們就從序列中去掉較小的值(即增大small),
* 相反，只需要增大big。
* 終止條件為：一直增加small到(1+sum)/2
*/
import java.util.ArrayList;
public
class No57FindContinuousSequence {public static void main(String[] args) {
No57FindContinuousSequence n = new No57FindContinuousSequence();
ArrayList<ArrayList<Integer>> lists = new ArrayList<ArrayList<Integer>>();
lists = n.FindContinuousSequence(15);
for (int i = 0; i < lists.size(); i++) {
System.out.println(lists.get(i));
}
}
public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum) {
ArrayList<ArrayList<Integer>> lists = new ArrayList<ArrayList<Integer>>();
if (sum < 3) {
return lists;
}
int small = 1;
int big = 2;
while (small < (sum + 1) / 2) {
int curSum = getSum(small, big);
if (curSum < sum) {
big++;
} else if (curSum == sum) {
ArrayList<Integer> list = new ArrayList<Integer>();
for (int i = small; i <= big; i++) {
list.add(i);
}
lists.add(list);
big++;
}else {
small++;
}
}
return lists;
}
private int getSum(int small, int big) {
int sum = 0;
for(int i = small; i <= big; i++) {
sum += i;
}
return sum;
}
}