代码随想录算法训练营第五十六天|1143.最长公共子序列 ● 1035.不相交的线 ● 53. 最大子序和 动态规划

1143. 最长公共子序列

int longestCommonSubsequence(char * text1, char * text2){
    int len1 = strlen(text1);
    int len2 = strlen(text2);
    int dp[len1+1][len2+1];
    for (int i = 0; i <= len1; i++)
    {
        for (int j = 0; j <= len2; j++)
        {
            dp[i][j] = 0;
        }
    }
    for (int i = 1; i <= len1; i++)
    {
        for (int j = 1; j <= len2; j++)
        {
            if (text1[i-1] == text2[j-1]) {
                dp[i][j] = dp[i-1][j-1] + 1;
            }else {
                dp[i][j] = fmax(dp[i-1][j], dp[i][j-1]);
            }
        }
    }   
    return dp[len1][len2];
}

1035. 不相交的线

与1143.最长公共子序列一样

int maxUncrossedLines(int* nums1, int nums1Size, int* nums2, int nums2Size){
    int len1 = nums1Size;
    int len2 = nums2Size;
    int dp[len1+1][len2+1];
    for (int i = 0; i <= len1; i++)
    {
        for (int j = 0; j <= len2; j++)
        {
            dp[i][j] = 0;
        }
    }
    for (int i = 1; i <= len1; i++)
    {
        for (int j = 1; j <= len2; j++)
        {
            if (nums1[i-1] == nums2[j-1]) {
                dp[i][j] = dp[i-1][j-1] + 1;
            }else {
                dp[i][j] = fmax(dp[i-1][j], dp[i][j-1]);
            }
        }
    }   
    return dp[len1][len2];
}

53. 最大子数组和

贪心算法：

int maxSubArray(int* nums, int numsSize){
    int result = INT_MIN;
    int count = 0;
    for (int i=0; i result) result = count;
        if (count < 0) count = 0;
    }
    return result;
}

动态规划：

int maxSubArray(int* nums, int numsSize){
    int dp[numsSize];
    for (int i = 0; i < numsSize; i++)
    {
        dp[i] = 0;
    }
    dp[0] = nums[0];
    int ans = nums[0];
    for (int j = 1; j < numsSize; j++)
    {
        // 比较的是当前数字大小和上次累加本次的大小
        dp[j] = fmax(nums[j], dp[j-1] + nums[j]);
        ans = fmax(dp[j], ans);
    }
    return ans;
}

整理一下

int maxSubArray(int* nums, int numsSize){
    int dp[numsSize];
    int ans = dp[0] = nums[0];
    for (int j = 1; j < numsSize; j++)
    {
        // 比较的是当前数字大小和上次累加本次的大小
        dp[j] = fmax(nums[j], dp[j-1] + nums[j]);
        if (dp[j] > ans) ans = dp[j];
    }
    return ans;
}