gobigandwin

public class Solution {

    // 从后往前倒序merge
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int index = m + n;
        while(m > 0 && n > 0) {
            if(nums1[m-1] >= nums2[n-1]) {
                nums1[--index] = nums1[--m];
            } else {
                nums1[--index] = nums2[--n];

gobigandwin

public class Solution {
    
    // O(n) time, O(n) space
    public int[] twoSum(int[] nums, int target) {
        int[] result = new int[2];
        if(nums == null || nums.length < 2) {
            return result;
        }
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        for(int i=0; i<nums.length; i++) {

gobigandwin

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
 
 // create a fake node first and then add two sorted list to the new list one by one

gobigandwin

// first, if str is null or white space
// second, sign is positive or negative
// third, compute result: no other letter than '0'-'9' and within Integer range
// fourth, make sure the range is between max and min Integer, pay attention to overflow!
public class Solution {
    public int myAtoi(String str) {
        if(str == null || str.length() == 0) {
            return 0;
        }
        str = str.trim();

gobigandwin

public class Solution {

    // new idea
    public boolean isValid(String s) {
        if(s == null || s.length() == 0) {
            return false;
        }
        Stack<Character> stack = new Stack<Character>();
        int len = s.length();
        for(int i=0; i<len; i++) {

gobigandwin

public class Solution {
    
    // We start with trimming.
    // If we see [0-9] we reset the number flags.
    // We can only see . if we didn't see e or ..
    // We can only see e if we didn't see e but we did see a number. We reset numberAfterE flag.
    // We can only see + and - in the beginning and after an e
    // any other character break the validation.
    // At the and it is only valid if there was at least 1 number and if we did see an e then a number after it as well.

gobigandwin

public class Solution {
    public boolean isPalindrome(String s) {
        if(s == null) {
            return false;
        }
        String str = s.trim().toLowerCase();
        int start = 0;
        int end = str.length() - 1;
        while(start < end) {
            char c1 = str.charAt(start);

gobigandwin

public class Solution {
    
    // method 1: fibonnaci f(n) = f(n-1) + f(n-2), 意思是当前的ways的数量是前一步ways数量以及前两步ways数量的和
    // time O(n) space O(1)
    public int climbStairs(int n) {
		if(n < 2) return n;
		int prev = 1;
		int cur = 1;
		for(int i=2; i<=n; i++) {
			int update = cur + prev;

gobigandwin

public class Solution {

    // Two Pointers, O(n^2)
    public ArrayList<ArrayList<Integer>> threeSum(int[] nums) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(nums == null || nums.length < 3) {
            return result;
        }
        Arrays.sort(nums);
        for(int i=0; i<nums.length-2; i++) {

gobigandwin

public class Solution {
    
    // solution 1: use constant space
    // 思路：把所有0都放在第一行或第一列
    public void setZeroes(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        if(m == 0 || n ==0) return;
        
        boolean hasZeroFirstRow = false;