LeetCode动态规划入门（专项打卡21天合集）
Posted 白鳯
tags:
篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了LeetCode动态规划入门（专项打卡21天合集）相关的知识，希望对你有一定的参考价值。
【LeetCode】动态规划入门（专项打卡21天合集）
下图为证
文章目录
Day1
斐波拉契数
class Solution
public int fib(int n)
int[] dp = new int[31];
dp[1] = 1;
for (int i = 2; i <= n; i++)
dp[i] = dp[i - 1] + dp[i - 2];
return dp[n];
第 N 个泰波那契数
class Solution
public int tribonacci(int n)
int[] dp = new int[38];
dp[1] = dp[2] = 1;
for (int i = 3; i <= n; i++)
dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3];
return dp[n];
Day2
爬楼梯
class Solution
public int climbStairs(int n)
if (n < 3) return n;
int cur = 2, pre = 1;
for (int i = 3; i <= n; i++)
int temp = cur;
cur += pre;
pre = temp;
return cur;
使用最小花费爬楼梯
class Solution
public int minCostClimbingStairs(int[] cost)
int n = cost.length;
int[] dp = new int[n];
dp[0] = cost[0];
dp[1] = cost[1];
for (int i = 2; i < n; i++)
dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
return Math.min(dp[n - 2], dp[n - 1]);
Day3
打家劫舍
class Solution
public int rob(int[] nums)
int n = nums.length;
int[] dp = new int[n + 1];
dp[1] = nums[0];
for (int i = 1; i < n; i++)
dp[i + 1] = Math.max(dp[i], dp[i - 1] + nums[i]);
return dp[n];
打家劫舍 II
class Solution
public int rob(int[] nums)
if (nums == null || nums.length == 0) return 0;
if (nums.length == 1) return nums[0];
int val1 = rob1(nums);
reverse(nums);
int val2 = rob1(nums);
return Math.max(val1, val2);
public int rob1(int[] nums)
int n = nums.length;
int[] dp = new int[n + 1];
dp[1] = nums[0];
for (int i = 1; i < n; i++)
dp[i + 1] = Math.max(dp[i], dp[i - 1] + nums[i]);
return dp[n - 1];
private void reverse(int[] nums)
int n = nums.length, i = 0;
while (i < n / 2)
int t = nums[i];
nums[i] = nums[n - i - 1];
nums[n - i - 1] = t;
i++;
删除并获得点数
class Solution
public int deleteAndEarn(int[] nums)
int n = 10000;
int[] mark = new int[n + 1];
for (int num : nums) mark[num] += num;
int[] dp = new int[n + 2];
dp[1] = mark[0];
for (int i = 1; i < n + 1; i++)
dp[i + 1] = Math.max(dp[i], dp[i - 1] + mark[i]);
return dp[n + 1];
Day4
跳跃游戏
前向遍历
class Solution
public boolean canJump(int[] nums)
int ri = 0;
for (int i = 0; i < nums.length; i++)
if (ri >= i)
ri = Math.max(ri, i + nums[i]);
else
return false;
return true;
后向遍历
class Solution
public boolean canJump(int[] nums)
if(nums == null || nums.length < 2) return true;
int le = nums.length - 1;
for(int i = nums.length - 2; i >= 0; i--)
if(i + nums[i] >= le) le = i;
return le == 0;
跳跃游戏 II
动态规划法
class Solution
public int jump(int[] nums)
int n = nums.length;
int[] dp = new int[n + 1];
for (int i = 0; i < nums.length; i++)
for (int j = 1; j <= nums[i] && i + j < n; j++)
dp[i + j] = dp[i + j] == 0 ? dp[i] + 1 : Math.min(dp[i + j], dp[i] + 1);
return dp[n - 1];
贪心法
class Solution
public int jump(int[] nums)
int steps = 0;
int reach = 0, nextreach = 0;
for (int i = 0; i < nums.length - 1; i++)
nextreach = Math.max(nextreach, i + nums[i]);
if (reach == i)
reach = nextreach;
steps++;
return steps;
Day5
最大子序和
class Solution
public int maxSubArray(int[] nums)
int res = nums[0], cur = nums[0];
for (int i = 1; i < nums.length; i++)
cur = cur > 0 ? cur + nums[i] : nums[i];
res = Math.max(res, cur);
return res;
环形子数组的最大和
class Solution
public int maxSubarraySumCircular(int[] nums)
int max, min, curMax, curMin, sum;
max = min = curMax = curMin = sum = nums[0];
for (int i = 1; i < nums.length; i++)
sum += nums[i];
curMax = curMax > 0 ? curMax + nums[i] : nums[i];
max = Math.max(max, curMax);
curMin = curMin < 0 ? curMin + nums[i] : nums[i];
min = Math.min(min, curMin);
return max < 0 ? max : Math.max(max, sum - min);
Day6
乘积最大子数组
class Solution
public int maxProduct(int[] nums)
int res, max, min;
res = max = min = nums[0];
for (int i = 1; i < nums.length; i++)
int mx = max, mn = min;
max = Math.max(mx * nums[i], Math.max(mn * nums[i], nums[i]));
min = Math.min(mx * nums[i], Math.min(mn * nums[i], nums[i]));
res = Math.max(res, max);
return res;
乘积为正数的最长子数组长度
class Solution
public int getMaxLen(int[] nums)
int res = 0, p = 0, q = 0;
for (int num : nums)
if (num == 0)
p = q = 0; //乘积为正负的长度
else
if (num > 0)
p++;
if (q > 0) q++;
else
int t = p;
p = q;
q = t;
q++;
if (p > 0) p++;
res = Math.max(res, p);
return res;
Day7
最佳观光组合
class Solution
public int maxScoreSightseeingPair(int[] values)
int res = 0, leftMax = 0;
for (int i = 0; i < values.length; i++)
res = Math.max(res, leftMax + values[i] - i);
leftMax = Math.max(leftMax, values[i] + i);
return res;
买卖股票的最佳时机
class Solution
public int maxProfit(int[] prices)
int profit = 0, minPrice = 10001;
for (int price : prices)
minPrice = Math.min(minPrice, price);
profit = Math.max(profit, price - minPrice);
return profit;
买卖股票的最佳时机 II
方法一：动态规划法
class leetcode专项 动态规划入门