实现一个函数：线段重叠

XX公司综合机试题第二题，以前有人讨论过，这里列出不同的算法

2.请实现一个函数：线段重叠; 输入多个一维线段,求出这些线段相交的所有区域(也用线段表示); 一条线段用两个值表示(x0,x1), 其中x1>x0;比如： 输入线段数组[(2,4),(1.5,6),(0.5,3.5),(5,7),(7.5,9)], 输出线段数组[(1.5,4),(5,6)]

public static Float[][] doTest(Float[][] array) {
	float[] temp = new float[array.length * 2];
	int[] count = new int[array.length * 2];
	for (int i = 0; i < array.length; i++) {
		for (int j = 0; j < array[i].length; j++) {
			temp[i * array[i].length + j] = array[i][j];
		}
	}
	Arrays.sort(temp);
	float x = 0f;
	float y = 0f;
	for (int i = 0; i < array.length; i++) {
		x = array[i][0];
		for (int j = 1; j < array[i].length; j++) {
			y = array[i][j];
			for (int k = 0; k < temp.length; k++) {
				if (temp[k] >= x && temp[k] < y)
					++count[k];
			}
		}
	}
	List<List<Float>> resultList = new LinkedList<List<Float>>();
	List<Float> list = null;
	int flag = 0;
	for (int i = 0; i < count.length; i++) {
		if (count[i] > 1 && flag == 0) {
			flag = 1;
			list = new LinkedList<Float>();
			list.add(temp[i]);
		} else if (count[i] > 1 && flag == 1) {
		} else if (count[i] == 1 && flag == 1) {
			flag = 0;
			list.add(temp[i]);
			resultList.add(list);
			list = null;
		}
	}
	if (resultList.size() > 0) {
		Float[][] result = new Float[resultList.size()][2];
		for (int i = 0; i < resultList.size(); i++) {
			list = resultList.get(i);
			for (int j = 0; j < list.size(); j++) {
				result[i][j] = list.get(j);
			}
		}
		return result;
	} else
		return null;
}