452. Minimum Number of Arrows to Burst Balloons

題目

There are some spherical balloons taped onto a flat wall that represents the XY-plane. The balloons are represented as a 2D integer array points where points[i] = [ $x_{start}$ , $x_{end}$ ] denotes a balloon whose horizontal diameter stretches between $x_{start}$ and $x_{end}$ . You do not know the exact y-coordinates of the balloons.

Arrows can be shot up directly vertically (in the positive y-direction) from different points along the x-axis. A balloon with $x_{start}$ and $x_{end}$ is burst by an arrow shot at x if $x_{start}$ <= x <= $x_{end}$ . There is no limit to the number of arrows that can be shot. A shot arrow keeps traveling up infinitely, bursting any balloons in its path.

Given the array points, return the minimum number of arrows that must be shot to burst all balloons.

Example 1

Input: points = [[10,16],[2,8],[1,6],[7,12]]
Output: 2
Explanation: The balloons can be burst by 2 arrows:
- Shoot an arrow at x = 6, bursting the balloons [2,8] and [1,6].
- Shoot an arrow at x = 11, bursting the balloons [10,16] and [7,12].

Example 2

Input: points = [[1,2],[3,4],[5,6],[7,8]]
Output: 4
Explanation: One arrow needs to be shot for each balloon for a total of 4 arrows.

Example 3

Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2
Explanation: The balloons can be burst by 2 arrows:
- Shoot an arrow at x = 2, bursting the balloons [1,2] and [2,3].
- Shoot an arrow at x = 4, bursting the balloons [3,4] and [4,5].

Constraints

1 <= points.length <= 105
points[i].length == 2
$-2^{31}$ <= $x_{start}$ < $x_{end}$ <= $2^{31} - 1$

題目要找最少做幾次可以把全部氣球刺破。

題目難易度

Medium

解題想法

題目給予每顆氣球 x 座標的起訖位置，也就是說如果氣球和氣球之間的起到訖有 1 個點重疊到，就可以一起處理。

我的作法是先將每顆氣球的起點位置從小到大排序，方便做比較。接下來拿每顆氣球的起點位置與前面的做比較(初始值起點和迄點給負無限是因為題目的 x 軸起點位置最小會是 $-2^{31}$ )。

假設每顆氣球的起點位置有在起點值和迄點之間，就代表可以一起刺破，同時要縮小起迄的位置，以確保可以一起刺破。若起點位置不在起點位置和迄點之間，代表是要多執行一次刺破的動作，並同時把起點位置和迄點位置替換成該氣球的路徑。

初試

Runtime: 352ms, 60%

Memory Usage: 75.6MB, 6.90%

/**
 * @param {number[][]} points
 * @return {number}
 */
var findMinArrowShots = function (points) {
  const x = points.sort((a, b) => a[0] - b[0]);

  let time = 0;
  let pS = -Infinity;
  let pE = -Infinity;
  for (let i = 0; i < x.length; i++) {
    const [start, end] = x[i];

    if (pS <= start && start <= pE) {
      pS = Math.max(pS, start);
      pE = Math.min(pE, end);
    } else {
      time++;
      pS = start;
      pE = end;
    }
  }

  return time;
};

題目​

題目難易度​

解題想法​

初試​

題目

題目難易度

解題想法

初試