【剑指offer-Java版】41和为s的两个数字VS和为s的连续正数序列

和为S的两个数字以及和为S的连续正数序列：输入是一个递增排序的数组，查找其中一对满足和为S的数，输出。如果有多对，输出其中一对

    public void FindNumbersWithSum(int nums[], int sum){
        if(nums == null) return;

        int indexL = 0;
        int indexR = nums.length - 1;
        while(indexL < indexR){
            if(nums[indexL] + nums[indexR] < sum) 
                indexL++; 
            else if(nums[indexL] + nums[indexR] > sum) 
                indexR--;
            else {
                System.out.println("nums["+indexL+"] + nums["+indexR+"] = "  + nums[indexL] + " + " + nums[indexR] + " = " + sum );
                indexL++;
                indexR--;
            }
        }
    }

    public void FindContinuousSequence(int sum){
        if(sum < 3) return;

        int mid = (1 + sum) / 2;
        int small = 1;
        int big = 2;
        int curSum = small;
        while(small < mid){ // 至少有两个数，而且是递增排序的，所以small最多为sum的一半
            if(curSum < sum){
                curSum += big;
                big++;
            }else if(curSum > sum){
                curSum -= small;
                small++;
            }else if(curSum == sum){
                for(int i=small; i" ");
                }
                System.out.println();
                curSum += big;
                big++;
            }
        }

    }

    }

测试代码：

    public class _Q41Test extends TestCase {

    _Q41 numWithSum = new _Q41();

    public void test(){
        int nums1[] = {1, 2, 4, 7, 11, 15};
        int sum1 = 15;
        numWithSum.FindNumbersWithSum(nums1, sum1);

        int nums2[] = {1, 2, 3, 4, 5, 6, 7, 8};
        int sum2 = 15;
        numWithSum.FindContinuousSequence(nums2, sum2);

        numWithSum.FindContinuousSequence(sum2);
    }
    }