[LeetCode] 801. Minimum Swaps To Make Sequences Increasing 最少交换使得序列递增
We have two integer sequences A and B of the same non-zero length.
We are allowed to swap elements A[i] and B[i]. Note that both elements are in the same index position in their respective sequences.
At the end of some number of swaps, A and B are both strictly increasing. (A sequence is strictly increasing if and only if A[0] < A[1] < A[2] < ... < A[A.length - 1].)
Given A and B, return the minimum number of swaps to make both sequences strictly increasing. It is guaranteed that the given input always makes it possible.
Example:
Input: A = [1,3,5,4], B = [1,2,3,7]
Output: 1
Explanation:
Swap A[3] and B[3]. Then the sequences are:
A = [1, 3, 5, 7] and B = [1, 2, 3, 4]
which are both strictly increasing.
Note:
A, Bare arrays with the same length, and that length will be in the range[1, 1000].A[i], B[i]are integer values in the range[0, 2000].
给两个长度相等的数组A和B,可在任意位置i交换A[i]和B[i]的值,使得数组A和B变成严格递增的数组,求最少需要交换的次数。
解法:dp
Java:
class Solution {
public int minSwap(int[] A, int[] B) {
int swapRecord = 1, fixRecord = 0;
for (int i = 1; i < A.length; i++) {
if (A[i - 1] >= B[i] || B[i - 1] >= A[i]) {
// In this case, the ith manipulation should be same as the i-1th manipulation
// fixRecord = fixRecord;
swapRecord++;
} else if (A[i - 1] >= A[i] || B[i - 1] >= B[i]) {
// In this case, the ith manipulation should be the opposite of the i-1th manipulation
int temp = swapRecord;
swapRecord = fixRecord + 1;
fixRecord = temp;
} else {
// Either swap or fix is OK. Let's keep the minimum one
int min = Math.min(swapRecord, fixRecord);
swapRecord = min + 1;
fixRecord = min;
}
}
return Math.min(swapRecord, fixRecord);
}
}
Python:
class Solution(object):
def minSwap(self, A, B):
"""
:type A: List[int]
:type B: List[int]
:rtype: int
"""
dp_no_swap, dp_swap = [0]*2, [1]*2
for i in xrange(1, len(A)):
dp_no_swap[i%2], dp_swap[i%2] = float("inf"), float("inf")
if A[i-1] < A[i] and B[i-1] < B[i]:
dp_no_swap[i%2] = min(dp_no_swap[i%2], dp_no_swap[(i-1)%2])
dp_swap[i%2] = min(dp_swap[i%2], dp_swap[(i-1)%2]+1)
if A[i-1] < B[i] and B[i-1] < A[i]:
dp_no_swap[i%2] = min(dp_no_swap[i%2], dp_swap[(i-1)%2])
dp_swap[i%2] = min(dp_swap[i%2], dp_no_swap[(i-1)%2]+1)
return min(dp_no_swap[(len(A)-1)%2], dp_swap[(len(A)-1)%2])
C++:
class Solution {
public:
int minSwap(vector<int>& A, vector<int>& B) {
int n = A.size();
vector<int> swap(n, n), noSwap(n, n);
swap[0] = 1; noSwap[0] = 0;
for (int i = 1; i < n; ++i) {
if (A[i] > A[i - 1] && B[i] > B[i - 1]) {
swap[i] = swap[i - 1] + 1;
noSwap[i] = noSwap[i - 1];
}
if (A[i] > B[i - 1] && B[i] > A[i - 1]) {
swap[i] = min(swap[i], noSwap[i - 1] + 1);
noSwap[i] = min(noSwap[i], swap[i - 1]);
}
}
return min(swap[n - 1], noSwap[n - 1]);
}
};
C++:
class Solution {
public:
int minSwap(vector<int>& A, vector<int>& B) {
int n1 = 0, s1 = 1, n = A.size();
for (int i = 1; i < n; ++i) {
int n2 = INT_MAX, s2 = INT_MAX;
if (A[i - 1] < A[i] && B[i - 1] < B[i]) {
n2 = min(n2, n1);
s2 = min(s2, s1 + 1);
}
if (A[i - 1] < B[i] && B[i - 1] < A[i]) {
n2 = min(n2, s1);
s2 = min(s2, n1 + 1);
}
n1 = n2;
s1 = s2;
}
return min(n1, s1);
}
};
类似题目:
Best Time to Buy and Sell Stock with Transaction Fee
All LeetCode Questions List 题目汇总
[LeetCode] 801. Minimum Swaps To Make Sequences Increasing 最少交换使得序列递增的更多相关文章
- LeetCode 801. Minimum Swaps To Make Sequences Increasing
原题链接在这里:https://leetcode.com/problems/minimum-swaps-to-make-sequences-increasing/ 题目: We have two in ...
- 801. Minimum Swaps To Make Sequences Increasing
We have two integer sequences A and B of the same non-zero length. We are allowed to swap elements A ...
- 【LeetCode】801. Minimum Swaps To Make Sequences Increasing 解题报告(Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 动态规划 参考资料 日期 题目地址:https:// ...
- 【leetcode】801. Minimum Swaps To Make Sequences Increasing
题目如下: We have two integer sequences A and B of the same non-zero length. We are allowed to swap elem ...
- 801. Minimum Swaps To Make Sequences Increasing 为使两个数组严格递增,所需要的最小交换次数
[抄题]: We have two integer sequences A and B of the same non-zero length. We are allowed to swap elem ...
- [LeetCode] Minimum Swaps To Make Sequences Increasing 使得序列递增的最小交换
We have two integer sequences A and B of the same non-zero length. We are allowed to swap elements A ...
- [Swift]LeetCode801. 使序列递增的最小交换次数 | Minimum Swaps To Make Sequences Increasing
We have two integer sequences A and B of the same non-zero length. We are allowed to swap elements A ...
- [LeetCode] 727. Minimum Window Subsequence 最小窗口序列
Given strings S and T, find the minimum (contiguous) substring W of S, so that T is a subsequence of ...
- 【LeetCode】Minimum Depth of Binary Tree 二叉树的最小深度 java
[LeetCode]Minimum Depth of Binary Tree Given a binary tree, find its minimum depth. The minimum dept ...
随机推荐
- navicat 远程连接服务器1130,1045问题报错处理
本人踩坑多次,一开始网上搜罗,解决办法大同小异,摸索了很久才全部解决完成,小小bug真磨人啊 首先,根据我的踩坑记录,navicat 1045和navicat 1130貌似属于同一种解决方案,都是修改 ...
- [NPM + React] Prepare a Custom React Hook to be Published as an npm Package
Before we publish our package, we want to make sure everything is set up correctly. We’ll cover vers ...
- React-Router常见API
React-Router是React项目中处理路由的库. 1. HashRouter 通过hashchange监听路由的变化,通过window.location.hash赋值触发监听的变化. 本质是一 ...
- 持续集成学习7 jenkins自动化代码构建
一.整体功能 1.触发上下游构建 2.我们在触发一个job的时候顺便丢一些参数过去,这些参数有可能是我这次编译过程中产生的一些地址,版本号或动态的一些东西丢到下游作为下游的构建参数 3.不同种类的视图
- NTSTATUS代码摘录
00000000 STATUS_SUCCESS00000000 STATUS_WAIT_000000001 STATUS_WAIT_100000002 STATUS_WAIT_200000003 ST ...
- js改变this指向
js中修改this的指向 方法整理 call,apply,bind 以上的三哥方法都是用来改变js中this的指向 call 使用方法:fun.call(thisArg[,arg1[, arg2[, ...
- java 数据库迁移工具 flyway
官方 https://github.com/flyway/flyway 简易demo https://github.com/deadzq/flyway-demo 主要在配置文件上做改动
- e3s10 网络管理
1. host上设置 iptables -t nat -A POSTROUTING -o eno1 -j MASQUERADE # https://www.unixtutorial.org/how-t ...
- 【luoguP2371】 [国家集训队]墨墨的等式
题目链接 考虑将所有的\(a_1x_1+a_2x_2+--+a_nx_n=B\)对\(a_1\)取模,那么所有可达到的B就分为了\(0\)~\(a_1-1\)类 如果对\(a_1\)取模为\(k\)的 ...
- 记一次CDH集群日志数据清理
背景 集群运行一段时间(大概一月多)后,cloudera manager管理界面出现爆红,爆红的组件有hdfs.zookeeper. 发现问题 点击详细内容查看,报日志空间不够的错误.初步判断是各个组 ...