第一题，汪仔换道具，主要思路就是先减去直接换的，然后排序，此时a=0，分为两种情况，b，c给a，以及c给b和a。设x解方程就可以求出表达式
方法一：
作者：千草幽幽
链接：https://www.nowcoder.com/discuss/500293?type=0&order=0&pos=7&page=1&channel=666&source_id=discuss_center_0
来源：牛客网
class Solution:
def numberofprize(self , a , b , c ):
# write code here
res = 0
res+= min(a,b,c)
a -= res
b -= res
c -= res
a,b,c = sorted([a,b,c])
return min(res + (2*b+c)//5,res + (b+c)//4)
方法二：
作者：neebie
链接：https://www.nowcoder.com/discuss/500271?type=all&order=time&pos=&page=1&channel=666&source_id=search_all
来源：牛客网
public int numberofprize (int a, int b, int c) {
// write code here
int left=min(a,b,c);
int right=max(a,b,c);
while (left<=right){
int mid=left+(right-left)/2;
if (check(mid,a,b,c)){
left=mid+1;
}else{
right=mid-1;
}
}
return right;
}
private boolean check(int mid, int a, int b, int c) {
long left=0;
if (a>=mid){
left+=(a-mid);
}else {
left-=(2*(mid-a));
}
if (b>=mid){
left+=(b-mid);
}else {
left-=(2*(mid-b));
}
if (c>mid){
left+=(c-mid);
}else {
left-=(2*(mid-c));
}
return left>=0;
}
private int min(int a, int b, int c) {
return Math.min(Math.min(a,b),c);
}
private int max(int a, int b, int c) {
return Math.max(Math.max(a,b),c);
}
第二题，建房子，思路是找房子间隔，大于t就+2，等于t就+1，最后考虑最左和最右，+2就可以了
class Solution:
def getHouses(self , t , xa ):
# write code here
res = 0
for i in range(0,len(xa),2):
xa[i],xa[i+1] = xa[i]-xa[i+1]/2,xa[i]+xa[i+1]/2
for i in range(len(xa)//2 -1):
if xa[2*i+2] - xa[2*i+1] == t:
res += 1
elif xa[2*i+2] - xa[2*i+1] > t:
res += 2
return res+2
方法二：
public int getHouses (int t, int[] xa) {
// write code here
if (xa==null||xa.length==0)
return 0;
double[][] xAndA=new double[xa.length/2][2];
int idx=0;
for (int i = 0; i < xa.length; i+=2) {
xAndA[idx][0]=(double)xa[i]-(double) xa[i+1]/2;
xAndA[idx][1]=(double)xa[i]+(double) xa[i+1]/2;
idx++;
}
int count=2;
for (int i = 0; i < xAndA.length-1; i++) {
if ((xAndA[i+1][0]-xAndA[i][1])==t){
count++;
}else if ((xAndA[i+1][0]-xAndA[i][1])>t){
count+=2;
}
}
return count;
}
第三题 求方差和最小的idx，这个题是原题= =有舍友阿里被问过。方法就是D(x) = E(X^2) - E(X)2,利用前缀和数组，可以很快计算出arr[i:j]的E(X)2和E(X^2).
class Solution:
def find_best_cut(self , arr ):
# write code here
n = len(arr)
prex = [0]
prex2 = [0]
for num in arr:
prex.append(prex[-1]+num)
prex2.append(prex2[-1]+num**2)
def var(i,j):
if i>=j:return 0
n = j-i+1
Ex = (prex[j+1]-prex[i])/n
Ex2 = (prex2[j+1]-prex2[i])/n
return Ex2-Ex**2
v = float('inf')
res = 0
for i in range(0,n):
temp = abs(var(0,i-1) + var(i,n-1))
if temp < v:
res = i
v = temp
return res
方法二：
public long getPasswordCount (String password) {
// write code here
if (password.length()==1)
return 9;
int i[]=new int[1];
Deque<Character> stack=new ArrayDeque<>();
backTrack(password.toCharArray(),0,i,stack);
return i[0];
}
private void backTrack(char[] chars, int start, int[] i, Deque<Character> stack) {
if (stack.size()==chars.length){
if (!checkEquals(stack,chars)){
i[0]++;
}
return;
}
if (start==0){
for (int j = 0; j <=9; j++) {
stack.addLast((char)('0'+(j-0)));
backTrack(chars,start+1,i,stack);
stack.pollLast();
}
}else{
int cur=((chars[start]-'0')+(stack.peekLast()-'0'));
if (cur%2==0){
stack.addLast((char)('0'+(cur/2-0)));
backTrack(chars,start+1,i,stack);
stack.pollLast();
}else{
for (int j = 0; j < 2; j++) {
stack.addLast((char)('0'+(cur/2+j-0)));
backTrack(chars,start+1,i,stack);
stack.pollLast();
}
}
}
}
private boolean checkEquals(Deque<Character> stack, char[] chars) {
int idx=0;
for (Character c:stack){
if (c!=chars[idx++])
return false;
}
return true;
}