leetcode DP

DP divide and approach分而治之

题号：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.

Apparently, this is a optimization problem, which can be usually solved by DP. So when it comes to DP, the first thing for us to figure out is the format of the sub problem(or the state of each sub problem). The format of the sub problem can be helpful when we are trying to come up with the recursive relation.

divide into small problem:  maxSubArray(int A[], int i), which means the maxSubArray for A[0:i ] which must has A[i] as the end element.

python

class Solution(object):
    def maxSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        dp = [0]*len(nums)
        dp[0] = nums[0]
        max_num = nums[0]
        for i in range(1,len(nums)):
            dp[i] = nums[i] + (dp[i-1] if dp[i-1] > 0 else 0)
            max_num = max(dp[i],max_num)
        return max_num

java

public int maxSubArray(int[] A) {
        int n = A.length;
        int[] dp = new int[n];//dp[i] means the maximum subarray ending with A[i];
        dp[0] = A[0];
        int max = dp[0];
        
        for(int i = 1; i < n; i++){
            dp[i] = A[i] + (dp[i - 1] > 0 ? dp[i - 1] : 0);
            max = Math.max(max, dp[i]);
        }
        
        return max;
}

其他答案：

python

class Solution:
    # @param A, a list of integers
    # @return an integer
    # 6:57
    def maxSubArray(self, A):
        if not A:
            return 0

        curSum = maxSum = A[0]
        for num in A[1:]:
            curSum = max(num, curSum + num)
            maxSum = max(maxSum, curSum)

        return maxSum

java

public static int maxSubArray(int[] A) {
    int maxSoFar=A[0], maxEndingHere=A[0];
    for (int i=1;i<A.length;++i){
        maxEndingHere= Math.max(maxEndingHere+A[i],A[i]);
        maxSoFar=Math.max(maxSoFar, maxEndingHere);    
    }
    return maxSoFar;
}

 169. Majority Element

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

python

class Solution(object):
    def majorityElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        dic = {}
        for item in nums:
            try:
                dic[item] +=  1
                if dic[item] > len(nums)//2:
                    return item
            except:
                dic[item] = 1
                if dic[item] > len(nums)//2:
                    return item

other:

python

重要，记住这个可以统计list array某元素个数的方式！！！！！！！

class Solution:
    def majorityElement(self, nums):
        majority_count = len(nums)//2
        for num in nums:
            count = sum(1 for elem in nums if elem == num)
            if count > majority_count:
                return num

class Solution:
    def majorityElement(self, nums):
        counts = collections.Counter(nums)
        return max(counts.keys(), key=counts.get)

研究一下collection这个包