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算法day1 两数之和两数相加冒泡排序快速排序

news 2025/7/7 2:22:56

两数之和

最简单的思维方式肯定是去凑两个数，两个数的和是目标值就ok。这里两遍for循环解决。

两数相加

敲了一晚上哈哈，结果超过int范围捏，难受捏。

public class Test2 {public static void main(String[] args) {
//        ListNode l1 = new ListNode(2);
//        l1.next = new ListNode(4);
//        l1.next.next = new ListNode(3);
//
//        ListNode l2 = new ListNode(5);
//        l2.next = new ListNode(6);
//        l2.next.next = new ListNode(4);ListNode l1 = new ListNode(5);ListNode l2 = new ListNode(5);ListNode listNode = new Test2().addTwoNumbers(l1, l2);System.out.println("l = " + listNode.val);System.out.println("l = " + listNode.next.val);
//        System.out.println("l = " + listNode.next.next.val);//        for (int i = 0; ; i++) {
//            ListNode l = listNode.next;
//            if (l.next != null) {
//                System.out.println("l = " + l.val);
//            }else return;
//        }}public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
//            Map map = new HashMap();
//            map.put("l1",l1.val+l1.next.val+l1.next.next.val)int num1 = l1.val;System.out.println("num1 = " + num1);int num2 = l2.val;ListNode l = l1.next;ListNode ll = l2.next;for (int i = 1; ; i++) {if ( l != null && l.next != null  ){num1 += l.val*Math.pow(10,i);l =  l.next;} else{if ( l != null  ){num1 +=  l.val*Math.pow(10,i);}System.out.println("num1 = " + num1);for (int j = 1; ; j++) {if (ll != null && ll.next != null){num2 += ll.val*Math.pow(10,j);ll =  ll.next;}else{if (ll != null ){num2 += ll.val*Math.pow(10,j);}System.out.println("num2 = " + num2);int temp = num1 +num2;System.out.println("temp = " + temp);int digit;int place = 1;  //位数ListNode lll = new ListNode();int count = 0;while (temp != 0) {temp = temp / 10;++count;}temp = num1 +num2;if (temp == 0)count=1;ListNode listNode = new ListNode();Map<String, ListNode> map =null;for (int i1 = 0; i1 < count; i1++) {digit = temp % 10;temp = temp / 10;System.out.println("Digit at place " + place + " is: " + digit);
//                            lll.val = digit;
//                             lll = lll.next;if (i1 == 0){lll.val = digit;if (count!=1){lll.next =   listNode;}}else if (i1 == 1){map = new HashMap<>();map.put("listNode"+i1,new ListNode());listNode.val = digit;if (count==2)return lll;listNode.next = map.get("listNode"+i1);System.out.println("map.get(\"listNode111\"+i1).val = " + map.get("listNode"+i1).val);} else{map.get("listNode"+(i1-1)).val = digit;if (i1==count-1)return lll;map.put("listNode"+i1,new ListNode());map.get("listNode"+(i1-1)).next = map.get("listNode"+i1);System.out.println("map.get(\"listNode222\"+i1).val = " + map.get("listNode"+i1).val);}place *= 10;}return lll;}}}}}}//两数求和
class ListNode {int val;ListNode next;ListNode() {}ListNode(int val) { this.val = val; }ListNode(int val, ListNode next) { this.val = val; this.next = next;}
}

冒泡排序

有这样一个数组：[ 6,1,2,4,6,2 ] 冒泡排序是将相邻两个元素比大小转换位置最终实现从小到大或者从大到小排序的过程。

首先关注第一个元素6：

6和1比转换位置

6和2比转换位置

6和4比转换位置

6和6比不动。

6和2比转换位置

再然后第二个元素到最后一个元素。这样每一个元素都找到他们对应的位置。

public static void sort1(int[] data){for (int i = 0; i < data.length; i++) {for (int j = 0; j < data.length - i - 1; j++) {if (data[j] > data[j + 1]){int k = data[j];data[j] = data[j + 1];data[j + 1] = k;}}}}