LeetCode Weekly Contest 34解题思路

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LeetCode Weekly Contest 34解题思路

详细代码可以fork下Github上leetcode项目，不定期更新。

赛题

本次周赛主要分为以下4道题：

• Leetcode 598. Range Addition II (5分)
• Leetcode 599. Minimum Index Sum of Two Lists (6分)
• Leetcode 565. Array Nesting (7分)
• Leetcode 600. Non-negative Integers without Consecutive Ones (9分)

Leetcode 598. Range Addition II (5分)

Problem:

Given an m * n matrix M initialized with all 0’s and several update operations. Operations are represented by a 2D array, and each operation is represented by an array with two positive integers a and b, which means Mi should be added by one for all 0 <= i < a and 0 <= j < b. You need to count and return the number of maximum integers in the matrix after performing all the operations.

Example:

Input:undefined m = 3, n = 3 operations = [2,2,3,3] Output: 4 Explanation:undefined Initially, M =undefined [0, 0, 0, 0, 0, 0, 0, 0, 0] After performing 2,2, M =undefined [1, 1, 0, 1, 1, 0, 0, 0, 0] After performing 3,3, M =undefined [2, 2, 1, 2, 2, 1, 1, 1, 1] So the maximum integer in M is 2, and there are four of it in M. So return 4.

Note:

• The range of m and n is 1,40000.
• The range of a is 1,m, and the range of b is 1,n.
• The range of operations size won’t exceed 10,000.

只说关键点，每次更新都是从(0,0)到(a,b)，所以只需要分别更新a，b为所有操作的最小值即可。

代码如下：

    public int maxCount(int m, int n, int[][] ops) {
        int x = m;
        int y = n;
        for (int[] op : ops){
            x = Math.min(x, op[0]);
            y = Math.min(y, op[1]);
        }
        return x * y;
    }

初始化为m和n，表明没有任何操作更新时，直接返回整个数组个数。

Leetcode 599. Minimum Index Sum of Two Lists (6分)

Problem:

Suppose Andy and Doris want to choose a restaurant for dinner, and they both have a list of favorite restaurants represented by strings. You need to help them find out their common interest with the least list index sum. If there is a choice tie between answers, output all of them with no order requirement. You could assume there always exists an answer.

Example 1:

Input: “Shogun”, “Tapioca Express”, “Burger King”, “KFC” Output: “Shogun” Explanation: The only restaurant they both like is “Shogun”.

Example 2:

Input: “Shogun”, “Tapioca Express”, “Burger King”, “KFC” Output: “Shogun” Explanation: The restaurant they both like and have the least index sum is “Shogun” with index sum 1 (0+1).

Note:

• The length of both lists will be in the range of 1, 1000.
• The length of strings in both lists will be in the range of 1, 30.
• The index is starting from 0 to the list length minus 1.
• No duplicates in both lists.

意思是说找共同喜欢的饭店，且index之和最小。一道map题，把list1中的key和下标记录到map中，接着如果在list2中存在同样的key，便更新下标和，同时记录最小sum，但最小sum的记录存在信息缺失。（index1 + index2 < list1.length，在扫描key时，可能会把list1中的key记录进来，而此key是不在list2中的），所以我的做法是把list1的长度加入进来，这样，index1 + index2 > list1.length， 排除这种情况。

代码如下：

    public String[] findRestaurant(String[] list1, String[] list2) {
        Map<String,Integer> map = new HashMap<>();
        for (int i = 0; i < list1.length; i++){
            map.put(list1[i], i);
        }
        int add = list1.length;
        int min = 1 << 30;
        for (int i = 0; i < list2.length; i++){
            if (map.containsKey(list2[i])){
                int value = map.get(list2[i]);
                map.put(list2[i], value + add + i);
                min = Math.min(min, value + add + i);
            }
        }
        List<String> ans = new ArrayList<>();
        for (String key : map.keySet()){
            int value = map.get(key);
            if (value == min){
                ans.add(key);
            }
        }
        return ans.toArray(new String[0]);
    }

Leetcode 565. Array Nesting (7分)

Problem:

A zero-indexed array A consisting of N different integers is given. The array contains all integers in the range 0, N - 1. Sets SK for 0 <= K < N are defined as follows: SK = { AK, A[AK], A[A[AK]], … }. Sets SK are finite for each K and should NOT contain duplicates. Write a function that given an array A consisting of N integers, return the size of the largest set SK for this array.

Example :

Input: A = 5,4,0,3,1,6,2 Output: 4 Explanation:undefined A0 = 5, A1 = 4, A2 = 0, A3 = 3, A4 = 1, A5 = 6, A6 = 2. One of the longest SK: S0 = {A0, A5, A6, A2} = {5, 6, 2, 0}

Note:

• N is an integer within the range 1, 20,000.
• The elements of A are all distinct.
• Each element of array A is an integer within the range 0, N-1.

union find, 尝试了递归，发现stack over flow，所以改成了迭代，需要注意，在搜索路径上，已经搜过的结点可以标记为-1，下次不需要再遍历。遍历所有可能的环，更新最大即可。

代码如下：

public int arrayNesting(int[] nums) {
        int size = 0;
        for (int i = 0; i < nums.length; i++){
            if (nums[i] == -1) continue;
            int next = nums[i];
            int step = 1;
            while (next != i){
                step++;
                int tmp = nums[next];
                nums[next] = -1;
                next = tmp;
            }
            size = Math.max(size, step);
        }
        return size;
    }

Leetcode 600. Non-negative Integers without Consecutive Ones (9分)

Problem:

Given a positive integer n, find the number of non-negative integers less than or equal to n, whose binary representations do NOT contain consecutive ones.

Example :

Input: 5 Output: 5 Explanation:undefined Here are the non-negative integers <= 5 with their corresponding binary representations: 0 : 0 1 : 1 2 : 10 3 : 11 4 : 100 5 : 101 Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule.

Note:

• 1 <= n <= 109

这道题较难，让我们求非连续1的所有可能的个数，当然关键问题在于要≤n\le n的情况。

≤n\le n困扰了我很久，但此问题其实可以归简为求：

给定指定位数n，求在给定位数下非连续1的个数。

例如：

n = 2;
00
01
10
11
ans = 3;

n = 3;
000
001
010
011
100
101
110
111
ans = 5

其实是一道排列组合题，排列组合题多半可以用递归式写出来，也就是说它是一道DP题，如何构建？

从 n-1 -> n 如何append？
假设f[n]为n位情况下，非连续1的个数，
则：
f[n] = "0"f[n-1] + "1"f[n-1];
分出一位，我们构建递归式，可以是0，也可以是1，所以有两种情况，但上述递归式有问题。

主要在于："1"f[n-1]，如果f[n-1]是非连续1的个数，而f[n-1]的解，并没有约束开头不能是"1"，所以
"1"f[n-1] = "10"f[n-2] + "11"f[n-2];
而"11"f[n-2]是非法的，所以：
f[n] = "0"f[n-1] + "10"f[n-2];
得：
f[n] = f[n-1] + f[n-2];

这样，我们可以求得任何位数的非连续1的个数。

但，该问题是让我们求≤n\le n的非连续1的个数，小于等于很关键，它的性质该如何利用？

例如：

num = 18

二进制表示如下：
18 = 0b10010

num = 1 0 0 1 0
pos = 4 3 2 1 0
开始利用f[n]，构建解。
当pos = 4 时，我们知道该位为1，那么它就存在两种情况：
1 X X X X  = num
0 X X X X  < num

= num 的暂且可以不考虑，而 < num 的这种情况，非连续1的个数为：f[4]
因为第一位是0，所以4位数的任何非连续1都是它的解。

接着看 = num 的情况，因为本来就是1，所以pos--；

pos = 3; 该位为9，由于 < num 的性质，不可能出现这位为1的情况，所以直接pos--;

pos = 2; 同上。

pos = 1;
1 0 0 1 X
1 0 0 0 X
同样存在两种情况，需要加上f[1];

上述情况num中并不存在连续的1，但给定的num也可能存在连续的1，如：
num = 12;

二进制表示如下：
12 = 0b1100;

num = 1 1 0 0;
pos = 3 2 1 0;

pos = 3时，可0可1
1 X X X
0 X X X f[3]

pos = 2时，可0可1？
不能为1，而只能为0！否则会出现连续1，所以只能是：
1 0 X X f[2]
直接跳出循环（因为f[2]已经包含了 1 0 X X的所有情况）。

代码如下：

public int findIntegers(int num) {
        int[] f = new int[32];
        f[0] = 1;
        f[1] = 2;
        for (int i = 2; i <= 31; i++) {
            f[i] = f[i - 1] + f[i - 2];
        }

        int pos = 31;
        while (pos >= 0 && ((num & (1 << pos)) == 0))
            pos--;
        if (pos == -1)
            return 1;

        int res = 0;
        int pre = 0;
        for (int i = pos; i >= 0; i--) {
            if (((num >> i) & 0x01) == 1) {
                res += f[i];
                if (pre == 1){
                    res --;
                    break;
                }
                pre = 1;
            }else{
                pre = 0;
            }
        }
        return res + 1;
    }