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AhkunTa
Arithmetic
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LeetCode
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452_javascript.js
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// 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y坐标并不重要,因此只要知道开始和结束的x坐标就足够了。开始坐标总是小于结束坐标。平面内最多存在104个气球。
// 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。
// Example:
// 输入:
// [[10,16], [2,8], [1,6], [7,12]]
// 输出:
// 2
// 解释:
// 对于该样例,我们可以在x = 6(射爆[2,8],[1,6]两个气球)和 x = 11(射爆另外两个气球)。
// 输入一个数值 x在 输入的数组内部 可以有 最大的值
/**
* @param {number[][]} points
* @return {number}
*/
// point[i] = [a,b]
// 题目意思就是 查询最大的不重叠空间
/**
* @param {number[][]} points
* @return {number}
*/
var findMinArrowShots = function (points) {
if (points.length == 0) return 0;
// 排序 用点的终点排序
points.sort((a, b) => a[1] - b[1]);
let end = points[0][1];
let count = 1;
for (let i = 0; i < points.length; i++) {
//若当前点的起点小于射击点,则代表当前一定包含射击点,可一箭射穿。
//若大于,则必须要重新射一直箭,并且把该点的终点作为区间判断依据
if (points[i][0] > end) {
count++;
end = points[i][1];
} else {
// 因为 根据终点大小排序 后面终点一定大于前面终点
// 此处可省略
// end = Math.min(end,points[i][1])
}
}
return count;
};
// 图形题解
//https://leetcode-cn.com/problems/minimum-number-of-arrows-to-burst-balloons/solution/yong-zui-shao-shu-liang-de-jian-yin-bao-qi-qiu-b-2/
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