[抄题]:

Given an array w of positive integers, where w[i] describes the weight of index i, write a function pickIndex which randomly picks an index in proportion to its weight.

Note:

  1. 1 <= w.length <= 10000
  2. 1 <= w[i] <= 10^5
  3. pickIndex will be called at most 10000 times.

Example 1:

Input:
["Solution","pickIndex"]
[[[1]],[]]
Output: [null,0]

Example 2:

Input:
["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
[[[1,3]],[],[],[],[],[]]
Output: [null,0,1,1,1,0]

[暴力解法]:

时间分析:

空间分析:

[优化后]:

时间分析:

空间分析:

[奇葩输出条件]:

[奇葩corner case]:

[思维问题]:

不知道怎么找index

[英文数据结构或算法,为什么不用别的数据结构或算法]:

二分查找index,最后还要在区间内讨论 区间的讨论需要带等号,从而返回一个整数

[一句话思路]:

把index和数字对应起来:在weightsum数组里搜索随机生成的weightsum,然后返回位置。

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

找index的二分要考虑相等的情况, 此时end = mid

[二刷]:

[三刷]:

[四刷]:

[五刷]:

[五分钟肉眼debug的结果]:

[总结]:

把index和数字对应起来:在weightsum数组里搜索随机生成的weightsum,然后返回位置。

[复杂度]:Time complexity: O(n) Space complexity: O(n)

[算法思想:迭代/递归/分治/贪心]:

[关键模板化代码]:

[其他解法]:

[Follow Up]:

[LC给出的题目变变变]:

[代码风格] :

[是否头一次写此类driver funcion的代码] :

[潜台词] :

class Solution {
Random random;
int[] w; public Solution(int[] w) {
//change the w to weightedsum
this.random = random;
for (int i = 1; i < w.length; i++) {
w[i] += w[i - 1];
}
this.w = w;
} public int pickIndex() {
//initilization
int start = 0; int end = w.length - 1;
int randomSum = random.nextInt(w[w.length - 1]) + 1; //search for a range
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (w[mid] == randomSum) {
end = mid;
}
if (w[mid] > randomSum) {
end = mid;
}if (w[mid] < randomSum) {
start = mid;}
} //search for the position
if (randomSum <= w[start]) return start;
else if (randomSum > w[end]) return end + 1;
else if (randomSum <= w[end]) return end; return -1;
}
} /**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(w);
* int param_1 = obj.pickIndex();
*/

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