每日一题--和为s的两个数||和为s的连续整数列

和为s的两个数：时间复杂度o(n)

package core;

/** 和为s的两个数字，输入数组为递增的
* Created by lxq on 2017/9/15.
*/
public class Problem5 {
public static void main(String[] args){
Problem5 problem5 = new Problem5();
int[] array = {1,2,4,6,7};
problem5.findNumberWithSum(array,11);
}

public boolean findNumberWithSum(int[] array,int sum){
boolean find = false;
if(array==null)
return find;
int number1 = 0; //结果
int number2 = 0; //结果
int start = 0;
int end = array.length-1;
while (start<end){
int curSum = array[start]+array[end];
if(curSum==sum){
number1 = array[start];
number2 = array[end];
find = true;
break;
}else if(curSum>sum){
end--;
}else {
start++;
}
}
System.out.println(number1);
System.out.println(number2);
return find;
}
}

和为s的连续整数列（至少有两个整数），则small最大不超过(s+1)/2，输出多个结果

与上面题目思路一样：设置small和big，big往后加，大的话就减去small,让small向后移动

package core;

/**输出所有和为s的连续整数序列（最少有两个数字）
* Created by lxq on 2017/9/15.
*/
public class Problem6 {
public static void main(String[] args){
Problem6 problem6 = new Problem6();
problem6.findContinueSequence(9);
}
public void findContinueSequence(int sum){
if(sum<3)
return;
int small = 1;
int big = 2;
int middle = (sum+1)/2;
int curSum = small + big;
//至少含有两个数字
while(small<middle){
if(curSum==sum) {
//打印从small到big的数字
printContinueNum(small, big);
}
while (curSum>sum&&small<middle){
//先将值减去，再变small值
curSum -= small;
small++;
if(curSum==sum)
printContinueNum(small,big);
}
//再增加big值，继续前面的过程，寻找更多的序列
//先变big值，再将值加上
big++;
curSum+=big;
}
}

private void printContinueNum(int small, int big) {
for(int i = small;i<=big;i++){
System.out.println(i+" ");
}
System.out.println();
}
}