面试题41：和为s的两个数VS和为s的连续正数数列

问题说明：

1.和为s的两个数问题是从一个排序的数组中找出和为s的两个数；

2.原题是找出一个即可，现在全部找出；

3.和为s的连续正数数列是给定一个数找出所有连续正数数列的和为s,例如s为9，（2,3,4）就是其中一组。

（一）和为s的两个数问题

public static int findNumbersWithSum(int[] sorted, int fromIndex, int toIndex, int sum){if(null == sorted || sorted.length == 0){throw new IllegalArgumentException();}// allow one element in sorted but at least two elment to make a pairif(fromIndex > toIndex){throw new IndexOutOfBoundsException("fromIndex : " + fromIndex + "toIndex : " + toIndex);}int countPair = 0;// return value int smallIndex = fromIndex;int bigIndex = toIndex;int currentSum = 0;while(smallIndex < bigIndex){currentSum = sorted[smallIndex] + sorted[bigIndex];if(currentSum > sum){bigIndex--;}if(currentSum < sum){smallIndex++;}if(currentSum == sum){int firstNotEqualSmall = smallIndex + 1;while(firstNotEqualSmall < bigIndex && (sorted[firstNotEqualSmall] == sorted[smallIndex])){firstNotEqualSmall++;}int firstNotEqualBig = bigIndex - 1;while(firstNotEqualBig > smallIndex && (sorted[firstNotEqualBig] == sorted[bigIndex])){firstNotEqualBig--;}//int equalSmallLen = firstNotEqualSmall - smallIndex;//int equalBigLen = bigIndex - firstNotEqualBig;// may be i == j so add condition by i < jfor(int i = smallIndex; i < firstNotEqualSmall; i++){for(int j = bigIndex; (i < j) && (j > firstNotEqualBig); j--){countPair++;System.out.println("equal sum = " + sum + " pair index is (" + i + "," + j + ")"                         + " and value is (" + sorted[i] + "," + sorted[j] + ")");}}// go on to find againsmallIndex = firstNotEqualSmall;bigIndex = firstNotEqualBig;}// end if ==}//end whilereturn countPair;}

（二）和为s的连续正数数列

public static void findContinousSequence(int sum){if(sum < 3){return ;}int small = 1;int big = 2;int currentSum = 3;int middle = (1 + sum) >> 1;while(small < middle){if(currentSum == sum){print(small, big);}while(currentSum > sum && small < middle){currentSum -= small;small++;if(currentSum == sum){print(small, big);}}// maybe currentSum < sum or small >= middlebig++;currentSum += big;}}