zoukankan      html  css  js  c++  java
  • 有关动态规划的LeetCode

    最长回文子串

    class Solution {
        public String longestPalindrome(String s) {
            if(s.equals("") || s == null) return "";
    int length = s.length();
    boolean[][] dp = new boolean[length][length];
    int start = 0;
    int end = 0;
    int res = 1;
    for(int i = length - 1; i >= 0; i --){
       
        for(int j = i ; j < length; j ++){
            if((j - i <= 2||dp[i+1][j-1])&&s.charAt(i)==s.charAt(j)){
                dp[i][j] = true;
            }
           if(dp[i][j]&&j - i + 1 > res){
               res = j - i + 1;
               start = i;
               end = j;
           }
         
        }
    }
    return s.substring(start,end+1);
        }
    }

    最长回文子序列

    class Solution {
        public int longestPalindromeSubseq(String s) {
    if(s == null || s.length()==0)
    return 0;
    int n = s.length();
    int[][] dp = new int[n][n];
    dp[n-1][n-1] = 1;
    for(int i = n-2; i >=0; i --){
        dp[i][i] = 1;
        for(int j = i+1; j < n; j ++){
            if(s.charAt(i)==s.charAt(j)){
             
                dp[i][j] = Math.max(dp[i][j],dp[i+1][j-1] + 2);
            }else{
                dp[i][j] = Math.max(dp[i+1][j],dp[i][j-1]);
            }
        }
    }
    return dp[0][n-1];
        }
    }

    最长上升子序列

    class Solution {
        public int lengthOfLIS(int[] nums) {
    if(nums.length == 0) return 0;
    int length = nums.length;
    int[] dp = new int[length];
    int res = 0;
    Arrays.fill(dp,1);
    for(int i = 0; i < length; i ++){
        for(int j = 0 ;j < i; j ++){
            if(nums[i] > nums[j]){
                dp[i] = Math.max(dp[i],dp[j]+1);
            }
        }
        res = Math.max(res,dp[i]);
    }
    return res;
        }
    }

    最长公共子序列

    class Solution {
        public int longestCommonSubsequence(String text1, String text2) {
    int m = text1.length();
    int n = text2.length();
    int[][] dp =new int[m+1][n+1];
    dp[0][0] = 0;
    for(int i = 1; i <= m; i ++){
        dp[i][0] = 0;
    }
    for(int i = 1; i <= n; i ++){
        dp[0][i] = 0;
    }
    for(int i = 1; i <= m; i++){
        for(int j = 1; j <= n; j ++){
            if(text1.charAt(i-1)==text2.charAt(j-1)){
                dp[i][j] = dp[i-1][j-1]+1;
            }else{
                dp[i][j] = Math.max(dp[i-1][j],dp[i][j-1]);
            }
        }
    }
    return dp[m][n];
        }
    }
  • 相关阅读:
    【题解】[USACO08DEC-Gold] Trick or Treat on the Farm
    【题解】[NOIP2015-TG] 信息传递
    【题解】[JLOI2011] 飞行路线
    平衡树
    斜率优化 dp 总结
    题解【P1833 樱花】
    题解【CodeForces 910A The Way to Home】
    三角恒等变换公式
    题解【洛谷 P1246 编码】
    生成函数(母函数)详解
  • 原文地址:https://www.cnblogs.com/moonyaoo/p/12843510.html
Copyright © 2011-2022 走看看