动态规划 53：Maximum Subarray，152：Maximum Subarray，266. Palindrome Permutation 回文全排列

 

题意：寻找子数组的和最大。

思路：设置dp数组来保存到第i位的最大和。

判断第i-1位的正负，若dp[i-1]<0 则 dp[i] = nums[i]; 若 dp[i-1] > 0 则 dp[i] = dp[i-1] +nums[i];

最后用 max_num = max(max_num, dp[i]); 来存储最大的和。

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int len = nums.size();
        if(len == 0) return 0;
        if(len == 1) return nums[0];
        vector<int> dp(len, 0);  //dp[i]: 以第i个元素为结尾的最大子序列和
        dp[0] = nums[0];
        int max_num = dp[0];
        for(int i=1; i<len; i++){
            if(dp[i-1] > 0)
                dp[i] = dp[i-1] +nums[i];
            else
                dp[i] = nums[i];
             
        }
        return max_num;
    }
};

题意：求子数组（数组里的数是连续）的最大乘积。

思路：因为数组里有负号，所以用两个数组dp_max 和 dp_min 分别保存到当前位置i的最大值和最小值。

状态转移方程：           

dp_max[i%2] = max( max(dp_max[(i-1)%2]*nums[i], nums[i]), dp_min[(i-1)%2]*nums[i]);
dp_min[i%2] = min(min(dp_min[(i-1)%2]*nums[i], nums[i]), dp_max[(i-1)%2]*nums[i]);

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int n = nums.size();
        if(n==0) return 0;
        int dp_max[2]={nums[0], 0}, dp_min[2]={nums[0], 0} ;
        int ans = nums[0];
        for(int i=1; i<n; i++){
            dp_max[i%2] = max( max(dp_max[(i-1)%2]*nums[i], nums[i]), dp_min[(i-1)%2]*nums[i]);
            dp_min[i%2] = min(min(dp_min[(i-1)%2]*nums[i], nums[i]), dp_max[(i-1)%2]*nums[i]);
            ans = max(ans, dp_max[i%2]);
        }
        
        return ans;
    }
};

266. Palindrome Permutation 回文全排列

Given a string, determine if a permutation of the string could form a palindrome.

Example 1:

Input: "code"
Output: false

Example 2:

Input: "aab"
Output: true

Example 3:

Input: "carerac"
Output: true

Hint:

1. Consider the palindromes of odd vs even length. What difference do you notice?
2. Count the frequency of each character.
3. If each character occurs even number of times, then it must be a palindrome. How about character which occurs odd number of times?