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 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.
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
有两个if时,为了防止两个if都不满足的情况,swap not_swap太小而搅屎棍干扰结果,初始值每次都设置成最大N
[思维问题]:
对dp很恐惧,没做过 不知道交换以后应该怎么检查,但是后续检查其实没有必要
[英文数据结构或算法,为什么不用别的数据结构或算法]:
数个数的dp需要新建数组
两个变量赋值相等,可以用连等号~
not_swap[i] = swap[i] = N;
[一句话思路]:
头一回做:递增可能不能换 能换可能不递增,所以需要把两步分开
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
有两个if时,为了防止两个if都不满足的情况,swap not_swap太小而搅屎棍干扰结果,初始值每次都设置成最大N
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[算法思想:递归/分治/贪心]:贪心
[关键模板化代码]:
坐标型:不存在前0位(没意义),第0位就能用 返回f[n - 1]
1- n位在循环中用,第0位直接在定义中用
swap[0] = 1;
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
class Solution {
public int minSwap(int[] A, int[] B) {
//ini: swap[1000], not_swap[1000]
int N = A.length;
int[] swap = new int[1000];
int[] not_swap = new int[1000];
swap[0] = 1;
not_swap[0] = 0; //for loop 1 < n
for (int i = 1; i < N; i++) {
swap[i] = N; not_swap[i] = N;
//compare normal or not
if (A[i - 1] < A[i] && B[i - 1] < B[i]) {
not_swap[i] = Math.min(not_swap[i], not_swap[i - 1]);
swap[i] = Math.min(swap[i], swap[i - 1] + 1);
}
//compare exchangeable or not
if (A[i - 1] < B[i] && B[i - 1] < A[i]) {
not_swap[i] = Math.min(not_swap[i], swap[i - 1]);
swap[i] = Math.min(swap[i], not_swap[i - 1] + 1);
}
} return Math.min(swap[N - 1], not_swap[N - 1]);
}
}
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