1.一维前缀和模板

解法:前缀和

java
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] array = new int[n + 1];
int m = scanner.nextInt();
for (int i = 1; i <= n; i++) {
array[i] = scanner.nextInt();
}
long[] dp=new long[n+1];
for (int i = 1; i <= n ; i++) {
dp[i] = dp[i - 1] + array[i];
}
for (int i = 0; i < m; i++) {
int left = scanner.nextInt();
int right = scanner.nextInt();
System.out.println(dp[right] - dp[left-1]);
}
}
2.二维前缀和模板

解法:前缀和

java
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m = scanner.nextInt();
int[][] array = new int[n + 1][m + 1];
int q = scanner.nextInt();
for (int i = 1; i <= n ; i++) {
for (int j = 1; j <= m ; j++) {
array[i][j] = scanner.nextInt();
}
}
long[][] dp = new long[n + 1][m + 1];
for (int i = 1; i <= n ; i++) {
for (int j = 1; j <= m ; j++) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + array[i][j];
}
}
for (int i = 0; i < q; i++) {
int x1 = scanner.nextInt();
int y1 = scanner.nextInt();
int x2 = scanner.nextInt();
int y2 = scanner.nextInt();
long ret = dp[x2][y2] - dp[x1 - 1][y2] - dp[x2][y1 - 1] + dp[x1 - 1][y1 - 1];
System.out.println(ret);
}
}
3.寻找数组的中心下标

解法:前缀和

java
public int pivotIndex(int[] nums) {
int len=nums.length;
int[] dp_left=new int[len];
int[] dp_right=new int[len];
for (int i = 1; i < len; i++) {
dp_left[i]=dp_left[i-1]+nums[i-1];
}
for (int i = len-2; i >=0; i--) {
dp_right[i]=dp_right[i+1]+nums[i+1];
}
for (int i = 0; i <len ; i++) {
if (dp_left[i]==dp_right[i]){
return i;
}
}
return -1;
}
4.除自身以外数组的乘积
238. 除了自身以外数组的乘积 - 力扣(LeetCode)

解法:前缀积 + 后缀积

java
public int[] productExceptSelf(int[] nums) {
int len=nums.length;
//定义前缀积数组
int[] dp_left=new int[len];
//定义后缀积数组
int[] dp_right=new int[len];
//处理边界情况
dp_left[0]=1;
dp_right[len-1]=1;
for (int i = 1; i <len ; i++) {
dp_left[i]=dp_left[i-1]*nums[i-1];
}
for (int i = len-2; i >=0 ; i--) {
dp_right[i]=dp_right[i+1]*nums[i+1];
}
int[] ret=new int[len];
for (int i = 0; i < len; i++) {
ret[i]=dp_left[i]*dp_right[i];
}
return ret;
}
5.和为k的子数组

解法:前缀和 + 哈希表

java
public int subarraySum(int[] nums, int k) {
HashMap<Integer,Integer> hashMap=new HashMap<>();
int sum=0;
int count=0;
//处理细节
hashMap.put(0,1);
for (int x:nums) {
sum+=x;
count+=hashMap.getOrDefault(sum-k,0);
hashMap.put(sum,hashMap.getOrDefault(sum,0)+1);
}
return count;
}
6.和可被K整除的子数组
974. 和可被 K 整除的子数组 - 力扣(LeetCode)

解法:前缀和

java
public int subarraysDivByK(int[] nums, int k) {
HashMap<Integer,Integer> hashMap=new HashMap<>();
int count=0;
int sum=0;
hashMap.put(0%k,1);
for (int x:nums) {
sum+=x;
int r=(sum%k+k)%k;
count+=hashMap.getOrDefault(r,0);
hashMap.put(r,hashMap.getOrDefault(r,0)+1);
}
return count;
}
7.连续数组

解法:前缀和

java
public int findMaxLength(int[] nums) {
HashMap<Integer,Integer> hashMap=new HashMap<>();
hashMap.put(0,-1);
int sum=0;
int len=0;
for (int i = 0; i < nums.length; i++) {
sum+=(nums[i]==0?-1:1);
if (hashMap.containsKey(sum)){
len=Math.max(len,i-hashMap.get(sum));
}else{
hashMap.put(sum,i);
}
}
return len;
}
8.矩阵区域和

解法:前缀和

java
public int[][] matrixBlockSum(int[][] mat, int k) {
int m=mat.length;
int n=mat[0].length;
int[][] dp=new int[m+1][n+1];
for (int i = 1; i <=m ; i++) {
for (int j = 1; j <=n ; j++) {
dp[i][j]=dp[i][j-1]+dp[i-1][j]-dp[i-1][j-1]+mat[i-1][j-1];
}
}
int[][] ans=new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int x1=Math.max(0,i-k)+1;
int y1=Math.max(0,j-k)+1;
int x2=Math.min(m-1,i+k)+1;
int y2=Math.min(n-1,j+k)+1;
ans[i][j]=dp[x2][y2]-dp[x1-1][y2]-dp[x2][y1-1]+dp[x1-1][y1-1];
}
}
return ans;
}