[leetcode] 题型整理之动态规划

动态规划属于技巧性比较强的题目，如果看到过原题的话，对解题很有帮助

55. Jump Game

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Determine if you are able to reach the last index.

For example:
A = [2,3,1,1,4], return true.

A = [3,2,1,0,4], return false.

不停扩展最远序号

代码较短，直接贴代码吧

public class Solution {
    public boolean canJump(int[] nums) {
        int length = nums.length;
        if (nums == null || length == 0) {
            return false;
        }
        int farthest = nums[0];
        for (int i = 0; i < length; i++) {
            if (i > farthest) {
                return false;
            }
            int x = i + nums[i];
            if (x > farthest) {
                farthest = x;
            }
            if (farthest >= length - 1) {
                return true;
            }
        }
        return farthest >= length - 1;
    }
}

 53. Maximum Subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.

代码当中，通过使用prev变量减少读取读取内存的次数。

public class Solution {
    public int maxSubArray(int[] nums) {
        int length = nums.length;
        if (length == 0) {
            return 0;
        }
        int[] rr = new int[length];
        rr[0] = nums[0];
        int result = nums[0];
        int prev = result;
        for (int i = 1; i < length; i++) {
            int x = prev > 0 ? prev : 0;
            prev = x + nums[i];
            rr[i] = prev;
            if (result < prev) {
                result = prev;
            }
        }
        return result;
    }
}

 63. Unique Paths II

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note: m and n will be at most 100.

从右下方向左上方获得各个位置到终点的路径数量。

public class Solution {
    int m;
    int n;
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        m = obstacleGrid.length;
        if (m == 0) {
            return 0;
        }
        n = obstacleGrid[0].length;
        if (n == 0) {
            return 0;
        }
        int[][] paths = new int[m][n];
        int mMinus = m - 1;
        int nMinus = n - 1;
        int x = paths[0][0];
        if (x == 1) {
            return 0;
        }
        x = obstacleGrid[mMinus][nMinus];
        if (x == 1) {
            return 0;
        } else {
            paths[mMinus][nMinus] = 1;
        }
        for (int i = n - 2; i >= 0; i--) {
            paths[mMinus][i] = obstacleGrid[mMinus][i] == 1 ? 0 : paths[mMinus][i + 1];
        }
        for (int i = m - 2; i >= 0; i--) {
            paths[i][nMinus] = obstacleGrid[i][nMinus] == 1 ? 0 : paths[i + 1][nMinus];
        }
        for (int i = m - 2; i >= 0; i--) {
            for (int j = n - 2; j >= 0; j--) {
                if (obstacleGrid[i][j] == 1) {
                    paths[i][j] = 0;
                } else {
                    paths[i][j] = paths[i + 1][j] + paths[i][j + 1];
                }
            }
        }
        return paths[0][0];
    }
}

 91. Decode Ways

A message containing letters from A-Z is being encoded to numbers using the following mapping:

'A' -> 1
'B' -> 2
...
'Z' -> 26

Given an encoded message containing digits, determine the total number of ways to decode it.

For example,
Given encoded message "12", it could be decoded as "AB" (1 2) or "L" (12).

The number of ways decoding "12" is 2.

public class Solution {
    public int numDecodings(String s) {
        int length = s.length();
        if (length == 0) {
            return 0;
        }
        int[] result = new int[length];
        if (s.charAt(0) == '0') {
            result[0] = 0;
        } else {
            result[0] = 1;
        }
        if (length == 1) {
            return result[0];
        }
        String string = s.substring(0, 2);
        int x = Integer.valueOf(string);
        if (s.charAt(1) == '0') {
            result[1] = 0;
            if (x <= 26 && x >= 10) {
                result[1]++;
            }
        } else {
            result[1] = result[0];
            if (x <= 26 && x >= 10) {
                result[1]++;
            }
        }
        for (int i = 2; i < length; i++) {
            char char0 = s.charAt(i);
            int r = 0;
            if (char0 != '0') {
                r += result[i - 1];
            }
            string = s.substring(i - 1, i + 1);
            x = Integer.valueOf(string);
            if (x <= 26 && x >= 10) {
                r += result[i - 2];
            }
            result[i] = r;
        }
        return result[length - 1];
    }
}

139. Word Break

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

public class Solution {
    public boolean wordBreak(String s, Set<String> wordDict) {
        s = " "+s;
        int l = s.length();
        boolean[] p = new boolean[l];
        p[0] = true;
        for(int i = 1; i < l; i++){
            for(int j = 0; j < i; j++){
                if(p[j] && wordDict.contains(s.substring(j+1,i+1))){
                    p[i] = true;
                    break;
                }
            }
        }
        if(p[l - 1]){
            return true;
        }
        else{
            return false;
        }
    }
}

142. Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Note: Do not modify the linked list.

Follow up:
Can you solve it without using extra space?

水中的鱼的详解

256. Paint House

There are a row of n houses, each house can be painted with one of the three colors: red, blue or green. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color.

The cost of painting each house with a certain color is represented by a n x 3 cost matrix. For example, costs[0][0] is the cost of painting house 0 with color red; costs[1][2] is the cost of painting house 1 with color green, and so on... Find the minimum cost to paint all houses.

Note:
All costs are positive integers.

public class Solution {
    public int minCost(int[][] costs) {
        int length = costs.length;
        if (length == 0) {
            return 0;
        }
        int[][] r = new int[length][3];
        r[0][0] = costs[0][0];
        r[0][1] = costs[0][1];
        r[0][2] = costs[0][2];
        int a0, a1, a2;
        for (int i = 1; i < length; i++) {
            a0 = r[i - 1][0];
            a1 = r[i - 1][1];
            a2 = r[i - 1][2];
            r[i][0] = Math.min(a1, a2) + costs[i][0];
            r[i][1] = Math.min(a0, a2) + costs[i][1];
            r[i][2] = Math.min(a0, a1) + costs[i][2];
        }
        int lengthMinus = length - 1;
        int x1 = r[lengthMinus][0];
        int x2 = r[lengthMinus][1];
        int x3 = r[lengthMinus][2];
        return x1 < x2 ? Math.min(x1, x3) : Math.min(x2, x3);
    }
}

 309. Best Time to Buy and Sell Stock with Cooldown

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions:

• You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
• After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)

Example:

prices = [1, 2, 3, 0, 2]
maxProfit = 3
transactions = [buy, sell, cooldown, buy, sell]

public class Solution {
    public int maxProfit(int[] prices) {
        final int length = prices.length;
        if (length < 2) {
            return 0;
        }
        int[] buys = new int[length];
        int[] sells = new int[length];
        buys[0] = 0 - prices[0];
        buys[1] = Math.max(-prices[0], -prices[1]);
        sells[0] = 0;
        sells[1] = Math.max(0, prices[1] - prices[0]);
        for (int i = 2; i < length; i++) {
            buys[i] = Math.max(buys[i - 1], sells[i - 2] - prices[i]);
            sells[i] = Math.max(sells[i - 1], buys[i - 1] + prices[i]);
        }
        return sells[length - 1];
    }
}

 325. Maximum Size Subarray Sum Equals k

Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn't one, return 0 instead.

Note:
The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range.

Example 1:

Given nums = [1, -1, 5, -2, 3], k = 3,
return 4. (because the subarray [1, -1, 5, -2] sums to 3 and is the longest)

Example 2:

Given nums = [-2, -1, 2, 1], k = 1,
return 2. (because the subarray [-1, 2] sums to 1 and is the longest)

Follow Up:
Can you do it in O(n) time?

public class Solution {
    public int maxSubArrayLen(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        int length = nums.length;
        int sum = 0;
        int result = 0;
        map.put(0, -1);
        for (int i = 0; i < length; i++) {
            sum += nums[i];
            if (!map.containsKey(sum)) {
                map.put(sum, i);
            }
            Integer r = map.get(sum - k);
            if (r != null) {
                result = Math.max(result, i - r);
            }
        }
        return result;
    }
}

 377. Combination Sum IV

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Example:

nums = [1, 2, 3]
target = 4

The possible combination ways are:
(1, 1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 3)
(2, 1, 1)
(2, 2)
(3, 1)

Note that different sequences are counted as different combinations.

Therefore the output is 7.

Follow up:
What if negative numbers are allowed in the given array?
How does it change the problem?
What limitation we need to add to the question to allow negative numbers?

public class Solution {
    public int combinationSum4(int[] nums, int target) {
        int[] rr = new int[target + 1];
        Arrays.fill(rr, 0);
        rr[0] = 1;
        Arrays.sort(nums);
        int length = nums.length;
        for (int i = 1; i <= target; i++) {
            for (int j =  0; j < length; j++) {
                int x = nums[j];
                int tt = i - x;
                if (tt < 0) {
                    break;
                }
                int prev = rr[tt];
                rr[i] += prev;
            }
        }
        return rr[target];
    }
}