力扣第62题:不同路径
class Solution {
public int uniquePaths(int m, int n) {
int[][] array = new int[m][n];
for(int i = 0;i<n;i++){
array[0][i] = 1;
}
for(int i = 0;i<m;i++){
array[i][0] = 1;
}
for(int i = 1;i<m;i++){
for(int j =1;j<n;j++ ){
array[i][j] = array[i - 1][j] + array[i] [j-1];
}
}
return array[m -1] [ n - 1];
}
}
力扣第64题:最小路径和
class Solution {
public int minPathSum(int[][] grid) {
int[][] array= new int[grid.length][grid[0].length];
array[0][0] = grid[0][0];
//行
for(int i = 1;i<grid[0].length;i++){
array[0][i] = array[0][i -1] + grid[0][i];
}
//列
for(int i = 1;i<grid.length;i++){
array[i][0] = array[i - 1][0] + grid[i][0];
}
for(int i = 1;i<grid.length;i++){
for(int j = 1;j<grid[0].length;j++){
array[i][j] = Math.min(array[i - 1][j] ,array[i][j - 1]) + grid[i][j];
}
}
return array[grid.length - 1][grid[0].length - 1];
}
}
力扣第287题:寻找重复数
class Solution {
public int findDuplicate(int[] nums) {
HashMap<Integer,Integer> map = new HashMap<>();
for(int num: nums){
if(map.containsKey(num)){
return num;
}
map.put(num,1);
}
return - 1;
}
}
力扣第108题:将有序数组转换为二叉搜索树
108. 将有序数组转换为二叉搜索树 - 力扣(LeetCode)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode sortedArrayToBST(int[] nums) {
return helper(nums,0,nums.length - 1);
}
private TreeNode helper(int nums[],int left,int right){
if(left > right){
return null;
}
int mid = (left + right) /2;
TreeNode s = new TreeNode(nums[mid]);
s.left = helper(nums,left,mid-1);
s.right = helper(nums,mid + 1,right);
return s;
}
}
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