1、字符串转换整数（atoi）

public int myAtoi(String s) {
    s=s.trim();
    if (s.length()==0) return 0;
    if (!Character.isDigit(s.charAt(0))&&s.charAt(0)!='+'&&s.charAt(0)!='-') return 0;
    boolean flag=true;
    if (s.charAt(0)=='+') {
        s=s.substring(1);
    }
    else if (s.charAt(0)=='-'){
        s=s.substring(1);
        flag=false;
    }
    long ans=0;
    for (int i=0;i<s.length();i++){
        if (!Character.isDigit(s.charAt(i))&&flag) return (int) ans;
        if (!Character.isDigit(s.charAt(i))&&!flag) return (int) -ans;
        int c=Integer.parseInt(s.charAt(i)+"");
        ans=ans*10+c;
        if (flag&&ans>=Integer.MAX_VALUE) return Integer.MAX_VALUE;
        if (!flag&&-ans<=Integer.MIN_VALUE) return Integer.MIN_VALUE;
    }
    return flag?(int)ans:(int)-ans;
}

2、回文数

public boolean isPalindrome(int x) {
    String s1=String.valueOf(x);
    StringBuilder s2=new StringBuilder();
    for (int i=s1.length()-1;i>=0;i--){
        s2.append(s1.charAt(i));
    }
    return s1.equals(s2.toString());
}

public boolean isPalindrome2(int x) {
    if (x<0||(x%10==0&&x!=0)) return false;
    int revertNum=0;
    while (x>revertNum){
        revertNum=revertNum*10+x%10;
        x/=10;
    }
    return x==revertNum||x==revertNum/10;
}

3、盛最多水的容器

public int maxArea(int[] height) {
    int ans=0,l=0,r=height.length-1;
    while (r>l){
        int area=Math.min(height[l],height[r])*(r-l);
        ans=Math.max(area,ans);
        if (height[r]>height[l]) ++l;
        else ++r;
    }
    return ans;
}