演算法導論（第三版）練習 10.1-1 ~ 10.1-7 兩個棧實現佇列 兩個佇列實現棧
阿新 • • 發佈：2018-11-17
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title></title> </head> <body> <script type="text/javascript"> let log = console.log.bind(console); let xView Code= [1, 2, 3]; x.isEmpty = function() { if (this.length == 0) { return true; } else { return false; } }; x.push(4); log("isEmpty", x.isEmpty()); // => isEmpty falselog(x); // => [1, 2, 3, 4] log(x.pop()); // => 4 log(x.pop()); // => 3 log(x.pop()); // => 2 log(x.pop()); // => 1 log(x.pop(), "??"); // => undefined ?? log("isEmpty", x.isEmpty()); // => isEmpty true</script> <script type="text/javascript"> // 利用陣列Q[0..n]（有效下標是0~n-1）實現一個最多容納n-1個元素的佇列 let queue = []; queue.size = 12; queue.tail = queue.head = 0; function enqueue(q, x) { let temp = (q.tail == q.size - 1) ? 0 : q.tail + 1; if (temp == q.head) { // 有效下標是12個的陣列，只能存11個數 throw new Error("overflow"); } else { q[q.tail] = x; q.tail = temp; } } function dequeue(q) { if (q.head == q.tail) { throw new Error("underflow"); } else { let x = q[q.head]; if (q.head == q.size - 1) { q.head = 1; } else { ++q.head; } return x; } } log(queue); for (let i = 0; i != queue.size - 1; ++i) { enqueue(queue, i); } log(queue); // => [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, size: 12, head: 0, tail: 11] for (let i = 0; i != queue.size - 1; ++i) { log(dequeue(queue)); } dequeue(queue); // => Uncaught Error: underflow </script> </body> </html>
10.1-1
跳
10.1-2
初始化：兩個指標分別指向0和x.length-1
push操作：首先檢查兩個是否相等，相等就丟擲overflow，不相等就賦值加1或者減1
10.1-3
跳
10.1-5
調個api。。
<script type="text/javascript"> let arr = [1, 2, 3, 4]; arr.push_back = function(x) { this.push(x); } arr.push_front = function(x) { this.unshift(x) } arr.pop_back = function() { return this.pop(); } arr.pop_front = function() { return this.shift(); } log(arr); // => [1, 2, 3, 4] arr.push_back(5); arr.push_front(0); log(arr); // => [0, 1, 2, 3, 4, 5] arr.pop_back(); arr.pop_front(); log(arr); // => [1, 2, 3, 4] </script>
10.1-6 & 10.1-7
不會做，不想做。。
大體思路參考：兩個棧實現佇列 兩個佇列實現棧