书脊 | 这青苔碧瓦堆, 俺曾睡风流觉, 将五十年兴亡看饱.

Simple Bank System

模拟一个银行系统,存钱取钱.

主要是判断一下各种可能情况.

class Bank {

    private long[] b;
    private int n;
    public Bank(long[] b) {
        this.n = b.length;
        this.b = new long[n + 1];
        for(int i = 0; i < n; i++){
            this.b[i+1] = b[i];
        }
    }
    private boolean exist(int a){
        return 1 <= a && a <= n;
    }
    public boolean transfer(int a1, int a2, long m) {
        if(exist(a1) && exist(a2) && b[a1] >= m){
            b[a1] -= m;
            b[a2] += m;
            return true;
        }
        else
            return false;
    }
    
    public boolean deposit(int a, long m) {
        if(!exist(a))
            return false;
        b[a] += m;
        return true;
    }
    
    public boolean withdraw(int a, long m) {
        if(!exist(a))
            return false;
        if(b[a] < m)
            return false;
        b[a] -= m;
        return true;
    }
}

/**
 * Your Bank object will be instantiated and called as such:
 * Bank obj = new Bank(balance);
 * boolean param_1 = obj.transfer(account1,account2,money);
 * boolean param_2 = obj.deposit(account,money);
 * boolean param_3 = obj.withdraw(account,money);
 */

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class Bank {

private long[] b;

private int n;

public Bank(long[] b) {

this.n = b.length;

this.b = new long[n + 1];

for(int i = 0; i < n; i++){

this.b[i+1] = b[i];

}

private boolean exist(int a){

return 1 <= a && a <= n;

}

public boolean transfer(int a1, int a2, long m) {

if(exist(a1) && exist(a2) && b[a1] >= m){

b[a1] -= m;

b[a2] += m;

return true;

}

else

return false;

}

public boolean deposit(int a, long m) {

if(!exist(a))

return false;

b[a] += m;

return true;

}

public boolean withdraw(int a, long m) {

if(!exist(a))

return false;

if(b[a] < m)

return false;

b[a] -= m;

return true;

}

/**

* Your Bank object will be instantiated and called as such:

* Bank obj = new Bank(balance);

* boolean param_1 = obj.transfer(account1,account2,money);

* boolean param_2 = obj.deposit(account,money);

* boolean param_3 = obj.withdraw(account,money);

*/

Date January 26, 2022
Category Leetcode
Comments No Comments

Smallest Index With Equal Value

给一个数组, 找最小的index, 使i mod 10 == nums[i].

用一个j表示一个从0->9的循环. 然后挨个找.

class Solution {
    public int smallestEqual(int[] nums) {
        int j = 0;
        for(int i = 0; i < nums.length; i++) {
            if(nums[i] == j)
                return i;
            j++;
            j %= 10;
        }
        return -1;
    }
}

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class Solution {

public int smallestEqual(int[] nums) {

int j = 0;

for(int i = 0; i < nums.length; i++) {

if(nums[i] == j)

return i;

j++;

j %= 10;

}

return -1;

}

Date January 26, 2022
Category Leetcode
Comments No Comments

Number of Pairs of Strings With Concatenation Equal to Target

给一个字符串组和一个target字符串, 问有多少个pair,组成target.

这题字符串组有重复, 所以先要counting一下, 然后切割target, 因为切割后的两个部分有可能相同, 相同的时候是C(n,n-1)个, 不同的时候是n* n-1个.

class Solution {
    public int numOfPairs(String[] nums, String target) {
        Map<String, Integer> count = new HashMap<>();
        for(String s : nums){
            count.put(s, count.getOrDefault(s, 0) + 1);
        }
        int res = 0;
        for(int i = 1; i < target.length(); i++) {
            String a = target.substring(0, i);
            String b = target.substring(i, target.length());
            if(count.containsKey(a) && count.containsKey(b)){
                if(a.equals(b)){ 
                    int n = count.get(a);
                    res += n * (n - 1);
                }else
                {
                    res += count.get(a) * count.get(b);
                }
            }
        }
        return res;
    }
}

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class Solution {

public int numOfPairs(String[] nums, String target) {

Map<String, Integer> count = new HashMap<>();

for(String s : nums){

count.put(s, count.getOrDefault(s, 0) + 1);

}

int res = 0;

for(int i = 1; i < target.length(); i++) {

String a = target.substring(0, i);

String b = target.substring(i, target.length());

if(count.containsKey(a) && count.containsKey(b)){

if(a.equals(b)){

int n = count.get(a);

res += n * (n - 1);

}else

{

res += count.get(a) * count.get(b);

}

return res;

}

Date January 26, 2022
Category Leetcode
Comments No Comments

Divide a String Into Groups of Size k

把一个字符串分割成k组, 如果不满k,就用fill字符补.

class Solution {
    public String[] divideString(String s, int k, char fill) {
        List<String> list = new ArrayList<>();
        int n = s.length();
        StringBuffer sb = new StringBuffer(s);
        int m = n % k == 0 ? 0 : k - (n%k);
        for(int i = 0; i < m; i++)
            sb.append(fill);
        String str = sb.toString();
        for(int i = 0; i < str.length(); i+=k){
            list.add(str.substring(i, i + k));
        }
        String[] strs = new String[list.size()];
        for(int i = 0; i < list.size(); i++)
            strs[i] = list.get(i);
        return strs;
    }
}

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class Solution {

public String[] divideString(String s, int k, char fill) {

List<String> list = new ArrayList<>();

int n = s.length();

StringBuffer sb = new StringBuffer(s);

int m = n % k == 0 ? 0 : k - (n%k);

for(int i = 0; i < m; i++)

sb.append(fill);

String str = sb.toString();

for(int i = 0; i < str.length(); i+=k){

list.add(str.substring(i, i + k));

}

String[] strs = new String[list.size()];

for(int i = 0; i < list.size(); i++)

strs[i] = list.get(i);

return strs;

}

Date January 24, 2022
Category Leetcode
Comments No Comments

Minimum Cost of Buying Candies With Discount

给一个数组, 是东西的价格, 可以买二送一, 求花费. 要求送的cost比买的要少.

class Solution {
    public int minimumCost(int[] cost) {
        Arrays.sort(cost);
        if(cost.length == 1)
            return cost[0];
        if(cost.length == 2)
            return cost[0] + cost[1];
        int res = 0;
        int i = cost.length - 1;
        while(i >= 0){
            if(i == 0){
                res += cost[i];
                break;
            }else if(i == 1){
                res += cost[i];
                res += cost[i - 1];
                break;
            }else{
                res += cost[i];
                res += cost[i - 1];
                i -= 3;
            }
        }
        return res;
    }
}

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class Solution {

public int minimumCost(int[] cost) {

Arrays.sort(cost);

if(cost.length == 1)

return cost[0];

if(cost.length == 2)

return cost[0] + cost[1];

int res = 0;

int i = cost.length - 1;

while(i >= 0){

if(i == 0){

res += cost[i];

break;

}else if(i == 1){

res += cost[i];

res += cost[i - 1];

break;

}else{

res += cost[i];

res += cost[i - 1];

i -= 3;

}

return res;

}

Date January 24, 2022
Category Leetcode
Comments No Comments

Count Elements With Strictly Smaller and Greater Elements

给一个数组, 问这个数组有多少个数字是至少有一个数字严格比它大还有一个严格比它小的.

class Solution {
    public int countElements(int[] nums) {
        int nMax = 0;
        int max = Integer.MIN_VALUE;
        int nMin = 0;
        int min = Integer.MAX_VALUE;
        for(int n : nums){
            max = Math.max(max, n);
            min = Math.min(min, n);
        }
        for(int n : nums){
            if(n == max)
                nMax++;
            if(n == min)
                nMin++;
        }
        if(nMax == nums.length)
            return 0;
        return nums.length - nMax - nMin;
    }
}

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class Solution {

public int countElements(int[] nums) {

int nMax = 0;

int max = Integer.MIN_VALUE;

int nMin = 0;

int min = Integer.MAX_VALUE;

for(int n : nums){

max = Math.max(max, n);

min = Math.min(min, n);

}

for(int n : nums){

if(n == max)

nMax++;

if(n == min)

nMin++;

}

if(nMax == nums.length)

return 0;

return nums.length - nMax - nMin;

}

Date January 24, 2022
Category Leetcode
Comments No Comments

Final Value of Variable After Performing Operations

class Solution {
    public int finalValueAfterOperations(String[] operations) {
        int res = 0;
        String a = "++X";
        String b = "X++";
        for(String o : operations)
        {
            if(o.equals(a) || o.equals(b))
                res++;
            else
                res--;
        }
        return res;
    }
}

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class Solution {

public int finalValueAfterOperations(String[] operations) {

int res = 0;

String a = "++X";

String b = "X++";

for(String o : operations)

{

if(o.equals(a) || o.equals(b))

res++;

else

res--;

}

return res;

}

Date January 11, 2022
Category Leetcode
Comments No Comments

Maximum Difference Between Increasing Elements

给一个数组, 找到两个数, i<j的差的最大值.

这题因为要求两个数有顺序, 所以只需要扫一次, 并且一直track最小值,就可以.

class Solution {
    public int maximumDifference(int[] nums) {
        int min = Integer.MAX_VALUE;
        int res = -1;
        for(int n : nums)
        {
            if(n <= min)
                min = n;
            else
                res = Math.max(n - min, res);
        }
        return res;
    }
}

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class Solution {

public int maximumDifference(int[] nums) {

int min = Integer.MAX_VALUE;

int res = -1;

for(int n : nums)

{

if(n <= min)

min = n;

else

res = Math.max(n - min, res);

}

return res;

}

Date January 11, 2022
Category Leetcode
Comments No Comments

Remove All Ones With Row and Column Flips

翻一个只有0和1的矩阵, 只能翻一个行或者一个列. 返回能不能翻成全1或者全0矩阵.

这题只需要判断第一行, 见到0不动, 见到1翻过来,这样第一行就是全0. 然后通过判断其他行是不是全1或者全0, 即可知道答案.

证: 如果第一行已经全0, 其他某行, 非全1或者全0, 那么这行通过行翻转, 肯定不能得到全1或者全0, 但是通过列翻转, 又破坏第一行的全0,故此.

class Solution {
    public boolean removeOnes(int[][] grid) {
        int n = grid.length;
        int m = grid[0].length;
        
        for(int i = 0; i < n; i++) {
            if(i == 0){ // if first row
                for(int j = 0; j < m; j++) {
                    if(grid[i][j] == 0){
                        for(int k = 0; k < n; k++){
                            flip(grid, k, j);
                        }
                    }
                }
            }
            else
            {
                for(int j = 1; j < m; j++) {
                    if(grid[i][j] != grid[i][j - 1])
                        return false;
                }
            }
        }
        return true;
    }
    
    public void flip(int[][] grid, int i, int j){
        if(grid[i][j] == 0)
            grid[i][j] = 1;
        else
            grid[i][j] = 0;
    }
   
}

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class Solution {

public boolean removeOnes(int[][] grid) {

int n = grid.length;

int m = grid[0].length;

for(int i = 0; i < n; i++) {

if(i == 0){ // if first row

for(int j = 0; j < m; j++) {

if(grid[i][j] == 0){

for(int k = 0; k < n; k++){

flip(grid, k, j);

}

else

{

for(int j = 1; j < m; j++) {

if(grid[i][j] != grid[i][j - 1])

return false;

}

return true;

}

public void flip(int[][] grid, int i, int j){

if(grid[i][j] == 0)

grid[i][j] = 1;

else

grid[i][j] = 0;

}

Date January 11, 2022
Category Leetcode
Comments No Comments

Capitalize the Title

给一个string, 是一段英文, 除了长度为1或者2的单词全部小写外, 大写每个单词的首字母.

class Solution {
    public String capitalizeTitle(String title) {
        String[] strs = title.split(" ");
        StringBuffer sb = new StringBuffer();
        for(String s : strs)
        {
            String ls = s.toLowerCase();
            if(ls.length() == 1 || ls.length() == 2)
                sb.append(ls);
            else
                sb.append(Character.toUpperCase(ls.charAt(0)) + ls.substring(1));
            sb.append(" ");
        }
        return sb.toString().trim();
    }
}

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class Solution {

public String capitalizeTitle(String title) {

String[] strs = title.split(" ");

StringBuffer sb = new StringBuffer();

for(String s : strs)

{

String ls = s.toLowerCase();

if(ls.length() == 1 || ls.length() == 2)

sb.append(ls);

else

sb.append(Character.toUpperCase(ls.charAt(0)) + ls.substring(1));

sb.append(" ");

}

return sb.toString().trim();

}

Date January 10, 2022
Category Leetcode
Comments No Comments

Simple Bank System

Smallest Index With Equal Value

Number of Pairs of Strings With Concatenation Equal to Target

Divide a String Into Groups of Size k

Minimum Cost of Buying Candies With Discount

Count Elements With Strictly Smaller and Greater Elements

Final Value of Variable After Performing Operations

Maximum Difference Between Increasing Elements

Remove All Ones With Row and Column Flips

Capitalize the Title

书脊

这青苔碧瓦堆, 俺曾睡风流觉, 将五十年兴亡看饱.

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