452. 用最少数量的箭引爆气球(难度:中等)
在二维空间中有许多球形的气球。对于每个气球,提供的输入是水平方向上,气球直径的开始和结束坐标。由于它是水平的,所以纵坐标并不重要,因此只要知道开始和结束的横坐标就足够了。开始坐标总是小于结束坐标。
一支弓箭可以沿着 x 轴从不同点完全垂直地射出。在坐标 x 处射出一支箭,若有一个气球的直径的开始和结束坐标为 xstart,xend, 且满足 xstart ≤ x ≤ xend,则该气球会被引爆。可以射出的弓箭的数量没有限制。 弓箭一旦被射出之后,可以无限地前进。我们想找到使得所有气球全部被引爆,所需的弓箭的最小数量。
给你一个数组 points ,其中 points [i] = [xstart,xend] ,返回引爆所有气球所必须射出的最小弓箭数。
解题思路
- 先将所有气球的坐标排序;
- 标记上一次炸裂的气球的尾端,如果其他气球的最前端在尾端之前,则会被同一只箭射中;
否则为另一只箭;
题解
public int findMinArrowShots(int[][] points) {
if (points.length == 0) return 0;
Arrays.sort(points, Comparator.comparingInt(o -> o[1]));
int minNumsOfArrows = 1;
int prevBurst = points[0][1];
for (int index = 1, pointsLength = points.length; index < pointsLength; index++) {
if (points[index][0] > prevBurst) {
++minNumsOfArrows;
prevBurst = points[index][1];
}
}
return minNumsOfArrows;
}
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测试
MinNumOfArrowsToBurstBalloons minNumOfArrowsToBurstBalloons = new MinNumOfArrowsToBurstBalloons();
@Test
public void test_case1() {
int[][] points = new int[][]{
{10, 16},
{2, 8},
{1, 6},
{7, 12}
};
Assertions.assertEquals(2, minNumOfArrowsToBurstBalloons.findMinArrowShots(points));
}
@Test
public void test_case2() {
int[][] points = new int[][]{
{1, 2},
{3, 4},
{5, 6},
{7, 8}
};
Assertions.assertEquals(4, minNumOfArrowsToBurstBalloons.findMinArrowShots(points));
}
@Test
public void test_case3() {
int[][] points = new int[][]{
{1, 2},
{2, 3},
{3, 4},
{4, 5}
};
Assertions.assertEquals(2, minNumOfArrowsToBurstBalloons.findMinArrowShots(points));
}
@Test
public void test_case4() {
int[][] points = new int[][]{
{1, 2},
};
Assertions.assertEquals(1, minNumOfArrowsToBurstBalloons.findMinArrowShots(points));
}
@Test
public void test_case5() {
int[][] points = new int[][]{
{2, 3},
{2, 3}
};
Assertions.assertEquals(1, minNumOfArrowsToBurstBalloons.findMinArrowShots(points));
}
@Test
public void test_case6() {
int[][] points = new int[][]{
{3, 9},
{7, 12},
{3, 8},
{6, 8},
{9, 10},
{2, 9},
{0, 9},
{3, 9},
{0, 6},
{2, 8}
};
Assertions.assertEquals(2, minNumOfArrowsToBurstBalloons.findMinArrowShots(points));
}
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