var a=1n; var b=1; a+b=2;()
时间: 2023-08-06 19:09:45 浏览: 61
这个表达式是错误的。在JavaScript中,BigInt类型的变量和number类型的变量不能直接相加,需要使用BigInt()函数将number类型的变量转换为BigInt类型。正确的表达式应该是:
```
var a = 1n;
var b = 1;
a + BigInt(b); // 返回结果为2n
```
或者
```
var a = BigInt(1);
var b = 1;
a + BigInt(b); // 返回结果为2n
```
需要注意的是,BigInt类型的变量在末尾需要加上字母n,否则会被认为是number类型的变量。
相关问题
function SuffixArray(str) { var s = str; var n = str.length; var sa = new Array(n), rank = new Array(n), height = new Array(n), x = new Array(n), y = new Array(n), m = 256, c = new Array(m); for (var i = 0; i < m; i++) c[i] = 0; for (var i = 0; i < n; i++) c[rank[i] = s.charCodeAt(i)]++; for (var i = 1; i < m; i++) c[i] += c[i - 1]; for (var i = n - 1; i >= 0; i--) sa[--c[rank[i]]] = i; for (var k = 1; k <= n; k <<= 1) { for (var i = n - k, p = 0; i < n; i++) x[p++] = i; for (var i = 0; i < n; i++) if (sa[i] >= k) x[p++] = sa[i] - k; for (var i = 0; i < m; i++) c[i] = 0; for (var i = 0; i < n; i++) c[rank[x[i]]]++; for (var i = 1; i < m; i++) c[i] += c[i - 1]; for (var i = n - 1; i >= 0; i--) sa[--c[rank[x[i]]]] = x[i]; for (var t = rank,rank = y,y = t,p = rank[sa[0]] = 0, i = 1; i < n; i++) { rank[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) ? p : ++p; } if (p >= n - 1) break; m = p + 1; } for (var k = i =0;i<n;++i){ if(rank[i]==0) continue; if(k) --k; var j=sa[rank[i]-1]; while(i+k<n && j+k<n && s.charAt(i+k)==s.charAt(j+k)) ++k; height[rank[i]]=k; } return { sa: sa, rank: rank, height: height }; } 这个函数创建后缀数组的能不能用DC3实现了,帮我修改为DC3算法可以吗
好的,这里是使用DC3算法实现后缀数组的代码:
function SuffixArray(str) {
var s = str;
var n = str.length;
var sa = new Array(n);
var rank = new Array(n);
var height = new Array(n);
var x = new Array(n);
var y = new Array(n);
var m = 256;
var c = new Array(m);
function sort_doubling(sa, rank, height, n, m) {
var tmp = new Array(n);
var i, j, p;
for (i = 0; i < m; ++i) c[i] = 0;
for (i = 0; i < n; ++i) c[x[i] = s.charCodeAt(i)]++;
for (i = 1; i < m; ++i) c[i] += c[i - 1];
for (i = n - 1; i >= 0; --i) sa[--c[x[i]]] = i;
for (j = p = 1; p < n; j <<= 1, m = p) {
for (p = 0, i = n - j; i < n; ++i) y[p++] = i;
for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
for (i = 0; i < m; ++i) c[i] = 0;
for (i = 0; i < n; ++i) c[x[y[i]]]++;
for (i = 1; i < m; ++i) c[i] += c[i - 1];
for (i = n - 1; i >= 0; --i) sa[--c[x[y[i]]]] = y[i];
for (tmp[sa[0]] = 0, i = 1, p = 1; i < n; ++i) {
if (x[sa[i]] == x[sa[i - 1]] && x[sa[i] + j] == x[sa[i - 1] + j]) {
tmp[sa[i]] = p - 1;
} else {
tmp[sa[i]] = p++;
}
}
for (i = 0; i < n; ++i) x[i] = tmp[i];
if (p >= n) break;
}
for (i = 0; i < n; ++i) rank[sa[i]] = i;
for (i = j = 0; i < n; ++i) {
if (rank[i]) {
if (j) --j;
var k = sa[rank[i] - 1];
while (i + j < n && k + j < n && s.charAt(i + j) == s.charAt(k + j)) ++j;
height[rank[i]] = j;
}
}
}
function sort_triples(Y, SA, n, K) {
var i, j;
if (K <= 1) {
for (i = 0; i < n; ++i) SA[i] = i;
} else {
var t = new Array(n + 3);
for (i = 0; i < n + 3; ++i) t[i] = 0;
for (i = 0; i < n; ++i) t[Y[i]]++;
for (i = 1; i < K; ++i) t[i] += t[i - 1];
for (i = 0; i < n; ++i) SA[--t[Y[i]]] = i;
for (j = 1; j < K; ++j) {
var p = 0;
for (i = n - j; i < n; ++i) y[p++] = i;
for (i = 0; i < n; ++i) if (SA[i] >= j) y[p++] = SA[i] - j;
for (i = 0; i < n + 3; ++i) t[i] = 0;
for (i = 0; i < n; ++i) t[x[i] = Y[i]]++;
for (i = 1; i < n + 3; ++i) t[i] += t[i - 1];
for (i = n - 1; i >= 0; --i) SA[--t[x[y[i]]]] = y[i];
var swap = x;
x = y;
y = swap;
p = 1;
x[SA[0]] = 0;
for (i = 1; i < n; ++i) {
var a = SA[i - 1], b = SA[i];
var cmp = ((Y[a] == Y[b]) && (Y[a + j] == Y[b + j]));
x[SA[i]] = cmp ? p - 1 : p++;
}
if (p >= n) break;
K = p;
}
}
}
function dc3(SA, s, n, K) {
var i, j, *y = new int[n + 3], * SA12 = new int[n / 3 + 1], * y12 = new int[n / 3 + 1], * SA0 = new int[n / 3], * y0 = new int[n / 3], c0, c1, c2;
s[n] = s[n + 1] = s[n + 2] = 0;
for (i = 0, j = 0; i < n; ++i) {
if (i % 3 != 0) y[j++] = i;
}
sort_triples(y, SA12, j, K);
for (i = 0; i < j; ++i) y12[SA12[i]] = i + 1;
for (i = 0, j = 0; i < n; ++i) {
if (i % 3 == 1) {
y0[j++] = i;
}
}
sort_doubling(SA0, y0, new int[j], j, K);
for (i = 0, j = 0, c0 = c1 = c2 = -1; i < n; ++i) {
if (i % 3 == 1) {
var p0 = SA0[j];
var p1 = (p0 < n - 1) ? y[p0 + 1] : -1;
var p2 = (p0 < n - 2) ? y[p0 + 2] : -1;
int cmp;
if ((cmp = (y[p0] == c0 ? y[p1] == c1 ? y[p2] == c2 ? 0 : 1 : 0 : 0)) == 1) {
SA[i] = p0;
++j;
} else {
SA[i] = n + SA12[y12[p0 / 3] + (cmp == 0 ? 0 : SA12.length)];
}
} else if (i % 3 == 2) {
var p0 = SA12[y[i / 3]];
var p1 = (p0 < n - 1) ? y[p0 + 1] : -1;
var p2 = (p0 < n - 2) ? y[p0 + 2] : -1;
int cmp;
if ((cmp = (y[p0] == c0 ? y[p1] == c1 ? y[p2] == c2 ? 0 : 1 : 0 : 0)) == 1) {
SA[i] = p0;
} else {
SA[i] = n + SA0[y0[p0 / 3] + (cmp == 0 ? 0 : SA0.length)];
}
}
}
delete[] y;
delete[] SA12;
delete[] y12;
delete[] SA0;
delete[] y0;
}
dc3(sa, s, n, m);
return {
sa: sa,
rank: rank,
height: height
};
}
实现一个函数从数组a中随机抽取取n个不重复元豪并返回经票var a =[1,2,3,4,5,6,7,8,9,10]; var n = 5: var b = getRandomArray(a,n); //实现getRandomArray
可以使用 Fisher-Yates 洗牌算法来从数组中随机取出 n 个不重复的元素。
具体实现如下:
```javascript
function getRandomArray(arr, n) {
const result = [];
const len = arr.length;
if (n >= len) {
// 如果要取出的元素个数不小于数组长度,则直接返回数组
return arr;
}
// Fisher-Yates 洗牌算法
for (let i = 0; i < n; i++) {
const randomIndex = Math.floor(Math.random() * (len - i));
result.push(arr[randomIndex]);
// 将已经选中的元素与最后一个元素交换位置
[arr[randomIndex], arr[len - 1 - i]] = [arr[len - 1 - i], arr[randomIndex]];
}
return result;
}
// 示例使用
const a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
const n = 5;
const b = getRandomArray(a, n);
console.log(b); // 输出随机抽取的 5 个不重复的元素
```
该函数的时间复杂度为 O(n),其中 n 为要取出的元素个数。
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