算法类解题思路

1.二叉树节点类
核心是递归，左右交叉的典型代码

    public static TreeNode invertTree(TreeNode root) {
        if(root == null){
            return null;
        }
        TreeNode right = invertTree(root.right);
        TreeNode left = invertTree(root.left);

        root.left = right;
        root.right = left;

        return root;
    }

2.字符串回文类型
核心是整体思想，将回文作为一个整体字符看待，下个字符相同时，右边扩张，字符不同时，两边扩张

    public static String longestPalindrome(String s){
        if(s == null || s.length() == 0){
            return "";
        }
        //保存起始位置
        char[] str = s.toCharArray();
        int maxLong = 0,bengin=0,last=0;
        for(int i=0;i<s.length();i++){
            //end代表重复字符串的最后一位
            int first = i;
            int end = i;
            //如果f(i) 的值与f(i+1)相等  代表中间部分重复的字符是同一个字符串
            while (end<s.length()-1 && str[first] == str[end+1]){
                end++;
            }
            //中间字符串向左右两边扩展，两边堆成
            while (first>0&&end<s.length()-1 && str[end+1] == str[first-1]){
                first--;
                end++;
            }
            //将end减去i就是回文的起始与结束位置
            if(end-first>maxLong){
                maxLong = end-first;
                bengin = first;
                last = end;
            }
        }
        return s.substring(bengin,last+1);
    }

3.动态规划的核心是动态转移方程（下一个f(i+1)依赖于f(i)）
最大乘积子数组

    public static int maxProduct(int[] nums) {
        int length = nums.length;
        int[] maxs = new int[length];
        int[] mins = new int[length];
        maxs[0] = mins[0] = nums[0];
        int maxResult = maxs[0];
        for (int i = 1; i < length; i++) {
            maxs[i] = Math.max(maxs[i - 1] * nums[i], Math.max(mins[i - 1] * nums[i], nums[i]));
            mins[i] = Math.min(mins[i - 1] * nums[i], Math.min(maxs[i - 1] * nums[i], nums[i]));
            maxResult = Math.max(maxResult, maxs[i]);
        }
        return maxResult;
    }