刷题总结
1，count数组
#include <iostream>
#include <unordered_map>
using namespace std;
// 输入： 11223344455
// 输出：{1:2, 2: 2, 3: 2, 4: 3, 5: 1}
void printNums(vector<int> nums) {
for(int i = 0; i < nums.size(); i++) {
cout << nums[i] << '\t';
}
}
vector<pair<int,int>> analysis(vector<int> nums) {
unordered_map<int, int> mp;
for(int i = 0; i < nums.size(); i++) {
if(mp.find(nums[i]) == mp.end()) {
mp.insert(pair(nums[i], 1));
} else {
mp[nums[i]] = mp.find(nums[i])->second + 1;
}
}
vector<pair<int, int>> result(mp.begin(), mp.end());
sort(result.begin(), result.end(), [](pair<int,int>&a, pair<int, int>&b) {
return a.second < b.second;
});
return result;
}
void printPair(vector<pair<int, int>> &nums) {
for(int i = 0; i < nums.size(); i++) {
cout << nums[i].first << ": " << nums[i].second << "\t";
}
}
int main() {
vector<int> nums = {7, 7, 2, 2, 3, 3, 4, 4 , 4, 5, 5};
// printNums(nums);
vector<pair<int, int>> result = analysis(nums);
printPair(result);
}2， leetcode 1. 两数之和
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> mp = new HashMap<>();
for(int i = 0; i < nums.length; i++) {
if(mp.containsKey(target - nums[i])) {
return new int[] {i, mp.get(target-nums[i])};
}
mp.put(nums[i], i);
}
return new int[2];
}
}
3，leetcode 4.寻找两个有序数组的中位数
解法一：
class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
// 合并两个数组
ArrayList<Integer> array = new ArrayList<>();
int i = 0, j = 0;
while(i < nums1.length || j < nums2.length) {
if(i < nums1.length && j < nums2.length && nums1[i] < nums2[j]) {
array.add(nums1[i++]);
} else if(j < nums2.length && i < nums1.length && nums1[i] >= nums2[j]) {
array.add(nums2[j++]);
}else if (j < nums2.length) {
array.add(nums2[j++]);
} else if (i < nums1.length) {
array.add(nums1[i++]);
}
}
int totalLength = nums1.length + nums2.length;
int middle = (totalLength - 1) / 2;
if(totalLength % 2 == 0) {
return (array.get(middle) + array.get(middle+1) )/ 2.0;
} else {
return array.get(middle);
}
}
}解法二：
class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int m = nums1.length;
int n = nums2.length;
if(m > n) {
return findMedianSortedArrays(nums2, nums1);
}
int c1 = 0, c2 = 0, lo = 0, hi = 2*m, LMAX1 = 0, LMAX2 = 0, RMIN1 = 0, RMIN2 = 0;
while(lo <= hi) {
c1 = (lo + hi) / 2;
c2 = m + n - c1;
LMAX1 = (c1 == 0) ? Integer.MIN_VALUE : nums1[(c1-1)/2];
LMAX2 = (c2 == 0) ? Integer.MIN_VALUE : nums2[(c2-1)/2];
RMIN1 = (c1 == 2*m) ? Integer.MAX_VALUE : nums1[c1/2];
RMIN2 = (c2 == 2*n) ? Integer.MAX_VALUE : nums2[c2/2];
if(LMAX1 > RMIN2) {
hi = c1 - 1;
} else if(LMAX2 > RMIN1) {
lo = c1 + 1;
} else {
break;
}
}
return (Math.max(LMAX1, LMAX2) + Math.min(RMIN1, RMIN2)) / 2.0;
}
}4，leetcode 11.盛水最多的容器
解法一（暴力）
class Solution {
public int maxArea(int[] height) {
int max = Integer.MIN_VALUE;
int len = height.length;
int tmp = 0;
for(int i = 0; i < len - 1; i++) {
for (int j = i + 1; j < len; j++) {
tmp = (j - i)*(Math.min(height[i], height[j]));
if(tmp > max) {
max = tmp;
}
}
}
return max;
}
}解法二（双指针）
class Solution {
public int maxArea(int[] height) {
int res = Integer.MIN_VALUE;
int i = 0, j = height.length - 1;
while(i < j) {
if(height[i] < height[j]){
res = Math.max(res, height[i]*(j - i));
i++;
} else {
res = Math.max(res, height[j]*(j - i));
j--;
}
}
return res;
}
}5, leetcode 15. 三数之和
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> result = new ArrayList();
int len = nums.length;
if(len < 3) return result;
Arrays.sort(nums);
for(int i = 0; i < len; i++) {
if(nums[i] > 0) break;
if(i > 0 && nums[i] == nums[i-1]) continue;
int L = i+1, R = len - 1;
while(L < R) {
int sum = nums[i] + nums[L] + nums[R];
if(sum == 0) {
result.add(Arrays.asList(nums[i], nums[L], nums[R]));
while(L < R && nums[L] == nums[L+1]) L++;
while(L < R && nums[R] == nums[R-1]) R--;
L++;
R--;
}
if(sum > 0) R--;
if(sum < 0) L++;
}
}
return result;
}
}6, leetcode 16. 最接近三数之和
class Solution {
public int threeSumClosest(int[] nums, int target) {
int len = nums.length;
if(len < 3) return 0;
int result = nums[0] + nums[1] + nums[2];
Arrays.sort(nums);
for(int i = 0; i < len; i++) {
int L = i + 1;
int R = len - 1;
while(L < R) {
int sum = nums[i] + nums[L] + nums[R];
result = Math.abs(result - target) > Math.abs(sum - target) ? sum : result;
if(sum > target) {
R--;
}else if(sum < target) {
L++;
}else {
return target;
}
}
}
return result;
}
}7, leetcode 18. 四数之和
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> result = new ArrayList();
int len = nums.length;
if(len < 4) return result;
Arrays.sort(nums);
for(int i = 0; i <= len - 4; i++) {
if(i > 0 && nums[i] == nums[i-1]) continue;
for(int j = i+1; j <= len - 3; j++) {
if(j > i+1 && nums[j] == nums[j-1]) continue;
int l = j+1, r = len - 1;
while(l < r) {
int sum = nums[i] + nums[j] + nums[l] + nums[r];
if(sum > target) {
r--;
}else if(sum < target) {
l++;
}else {
result.add(Arrays.asList(nums[i], nums[j], nums[l], nums[r]));
while(nums[l]==nums[l+1]) l++;
while(nums[r] == nums[r-1]) r--;
l++;
r--;
}
}
}
}
return result;
}
}8. leetcode 26. 删除排序数组中的重复项
解法一：（暴力移动法）
class Solution {
public int removeDuplicates(int[] nums) {
//判断重复 记录位置 删除
int end = nums.length, i = 0;
while(i < end) {
if(i > 0 && nums[i] == nums[i-1]) {
//移动
for(int j = i+1; j < end; j++) {
nums[j-1] = nums[j];
}
end--;
}else {
i++;
}
}
return end;
}
}解法二：（双指针法）
class Solution {
public int removeDuplicates(int[] nums) {
int len = nums.length;
if(len == 0) return 0;
else if(len == 1) return 1;
else {
int i = 0;
for(int j = 1; j < len; j++) {
if(nums[i] != nums[j]) {
i++;
nums[i] = nums[j];
}
}
return i+1;
}
}
}9. leetcode 3. 无重复字符的最长字串
解法一：
class Solution {
public int lengthOfLongestSubstring(String s) {
int len = s.length();
int i = 0;
int result = 0;
while(i < len) {
Set<String> set= new HashSet();
int j = 0;
for(j = i; j < len; j++) {
if(!set.contains(s.charAt(j)+"")) {
result = Math.max(result, j-i+1);
}else {
break;
}
set.add(s.charAt(j)+"");
}
i++;
}
return result;
}
}解法二：（滑动窗口）
class Solution {
public int lengthOfLongestSubstring(String s) {
int LeftIndex = 0;
int result = 0;
for(int j = 0; j < s.length(); j++) {
for(int CurIndex = LeftIndex; CurIndex < j; CurIndex++) {
if(s.charAt(CurIndex) == s.charAt(j)) {
result = Math.max(result, j - LeftIndex);
LeftIndex = CurIndex + 1;
break;
}
}
}
return Math.max(s.length()-LeftIndex, result);
}
}10. leetcode 27. 移除元素
解法一：
class Solution {
public int removeElement(int[] nums, int val) {
int i = 0;
for(int j = 0; j < nums.length; j++) {
if(nums[j] != val) {
nums[i] = nums[j];
i++;
}
}
return i;
}
}解法二：
class Solution {
public int removeElement(int[] nums, int val) {
int len = nums.length;
int i = 0;
while(i < len) {
if(nums[i] == val) {
nums[i] = nums[len-1];
len--;
}else {
i++;
}
}
return len;
}
}11. leetcode 6. Z字形变换
解法一：
class Solution {
public String convert(String s, int numRows) {
int len = s.length();
//根据s的长度 和numRows来分配初始数组的大小
StringBuilder sb = new StringBuilder();
if(len == 0) return sb.toString();
if(numRows == 1) return s;
int numCols = (len / (2*numRows - 2) + 1) * (numRows-1);
char[][] result = new char[numRows][numCols];
int i = 0, row = 0, col = 0;
while(i < len) {
for(int j = 0; j < numRows && i < len; j++) {
result[j][col] = s.charAt(i++);
}
col++;
for(int j = numRows-2; j >= 1 && i < len; j--) {
result[j][col++] = s.charAt(i++);
}
}
for(i = 0; i < result.length; i++) {
for(int j = 0; j < result[0].length; j++) {
if(result[i][j] != '\0') {
sb.append(result[i][j]);
}
}
}
return sb.toString();
}
}解法二：
class Solution {
public String convert(String s, int numRows) {
int len = s.length();
numRows = Math.min(numRows, len);
if(numRows == 1) return s;
//构造stringbuilder
List<StringBuilder> sb = new ArrayList<>();
for(int i = 0; i < numRows; i++) {
sb.add(new StringBuilder());
}
boolean goDown = false;
int row = 0;
for(int i = 0; i < len; i++) {
sb.get(row).append(s.charAt(i));
if(row == 0 || row == numRows - 1) goDown = !goDown;
row += goDown ? 1 : -1;
}
StringBuilder res = new StringBuilder();
for(int i = 0; i < sb.size(); i++) {
res.append(sb.get(i));
}
return res.toString();
}
}12. leetcode 5. 最长回文子串
解法一：(暴力)
class Solution {
public String longestPalindrome(String s) {
if(s.length() < 2) return s;
String res = s.substring(0, 1);
int maxLen = 1;
for(int i = 0; i < s.length() - 1; i++) {
for(int j = i + 1; j < s.length(); j++) {
if(j-i+1 > maxLen && valid(i, j, s)) {
maxLen = j - i + 1;
res = s.substring(i, j+1);
}
}
}
return res;
}
public boolean valid(int i, int j, String s) {
char[] ss = s.toCharArray();
while(i <= j) {
if(ss[i++] != ss[j--]) {
return false;
}
}
return true;
}
}解法二：（中心扩散法）
class Solution {
public String longestPalindrome(String s) {
int len = s.length();
if(len < 2) return s;
String res = s.substring(0, 1);
int maxLen = 1;
for(int i = 0; i < len-1; i++) {
String s1 = centerS(s, i, i);
String s2 = centerS(s, i, i+1);
String maxS = s1.length() > s2.length() ? s1 : s2;
if(maxS.length() > maxLen) {
maxLen = maxS.length();
res = maxS;
}
}
return res;
}
public String centerS(String s, int l, int r) {
int len = s.length();
while(l >= 0 && r < len) {
if(s.charAt(l) == s.charAt(r)) {
l--;
r++;
} else {
break;
}
}
return s.substring(l+1, r);
}
}解法三：（动态规划）
class Solution {
public String longestPalindrome(String s) {
int len = s.length();
if(len < 2) return s;
boolean[][] dp = new boolean[len][len];
String res = s.substring(0, 1);
int maxLen = 1;
for(int r = 1; r < len; r++) {
for(int l = 0; l < r ; l++) {
if(s.charAt(l) == s.charAt(r) && (r - l <= 2 || dp[l+1][r-1])) {
dp[l][r] = true;
if(r - l + 1 > maxLen) {
maxLen = r - l + 1;
res = s.substring(l, r+1);
}
}
}
}
return res;
}
}13. leetcode 102. 二叉树的层次遍历
解法一：（递归）
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public static List<List<Integer>> res;
public List<List<Integer>> levelOrder(TreeNode root) {
res = new ArrayList<>();
if(root == null) return res;
helper(root, 0);
return res;
}
public void helper(TreeNode root, int level) {
if(res.size() == level) {
res.add(new ArrayList());
}
res.get(level).add(root.val);
if(root.left != null) {
helper(root.left, level+1);
}
if(root.right != null) {
helper(root.right, level+1);
}
}
}解法二：（迭代）
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if(root == null) return res;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()) {
int _size = queue.size();
List<Integer> tmp = new ArrayList<>();
for(int i = 0; i < _size; i++) {
TreeNode node = queue.remove();
tmp.add(node.val);
if(node.left != null) {
queue.offer(node.left);
}
if(node.right != null) {
queue.offer(node.right);
}
}
res.add(tmp);
}
return res;
}
}14. leetcode 103. 二叉树的锯齿形层次遍历
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if(root == null) return res;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()) {
int _size = queue.size();
List<Integer> tmp = new ArrayList<>();
for(int i = 0; i < _size; i++) {
TreeNode node = queue.remove();
tmp.add(node.val);
if(node.left != null){
queue.offer(node.left);
}
if(node.right != null) {
queue.offer(node.right);
}
}
if(res.size()%2 == 0) {
res.add(tmp);
} else {
Collections.reverse(tmp);
res.add(tmp);
}
}
return res;
}
}15, leetcode 100.相同的树
解法一：（递归）
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if(p == null && q == null) return true;
else if(p == null || q == null) return false;
else {
if(p.val != q.val) return false;
}
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
}