publicstaticintscoreOfString(String s){int num=0;for(int i=1;i<s.length();i++){num+=Math.abs(s.charAt(i)-s.charAt(i-1));}returnnum;} Runtime:1 ms, faster than100.00%of Java online submissions for Score of a String. Memory Usage:42 MB, less than50.00%of Java online submissio...
Score of a String/main.scala Original file line numberDiff line numberDiff line change @@ -0,0 +1,9 @@ object Solution { def scoreOfString(s: String): Int = { var sum: Int = 0 for (i <- 1 to s.length-1) { sum += (s.charAt(i-1) - s.charAt(i)).abs } return sum...
Given a balanced parentheses string s, return the score of the string. The score of a balanced parentheses string is based on the following rule: "()" has score 1. AB has score A + B, where A and B are balanced parentheses strings. (A) has score 2 * A, where A is a balanced ...
Can you solve this real interview question? Maximum Score After Splitting a String - Given a string s of zeros and ones, return the maximum score after splitting the string into two non-empty substrings (i.e. left substring and right substring). The sco
Leetcode 856. Score of Parentheses 括号得分(栈) 题目描述 字符串S包含平衡的括号(即左右必定匹配),使用下面的规则计算得分 () 得1分 AB 得A+B的分,比如()()得2分 (A) 得2A分, 比如(()())得2(1+1)分 测试样例 详细分析 简而言之,遇到右括号就一直出栈并累加到一个值直到遇到左括号,这个累加值...
Runtime: 16 ms, faster than 94.77% of Python online submissions for Maximum Score After Splitting a String. Memory Usage: 13.4 MB, less than 65.70% of Python online submissions for Maximum Score After Splitting a String. 1 2 原题链接:https://leetcode.com/problems/maximum-score-after-split...
package leetcode func scoreOfParentheses(S string) int { res, stack, top, temp := 0, []int{}, -1, 0 for _, s := range S { if s == '(' { stack = append(stack, -1) top++ } else { temp = 0 for stack[top] != -1 { temp += stack[top] stack = stack[:len(stack...
LeetCode 32. Longest Valid Parentheses Given astringcontaining just the characters '(' and ')', find the lengthofthe longest valid (well-formed)parenthesessubstring. Example 1: Example 2: 解法一:用栈stack 解法二:动态规划DP 【LeetCode】856. Score of Parentheses 解题报告(Python & C++) ...
动态规划:leetcode 1039: Minimum Score Triangulation of Polygon,376:Wiggle Subsequence,678:Valid Parenthesis String 思路:动态规划dp[i][j]:从第i个点到第j个点的最小值。dp[i][j] = min(dp[i][j], dp[i][k]+dp[k][j]+A[i]*A[j]*A[k])...
https://github.com/grandyang/leetcode/issues/856 类似题目: Valid Parentheses 参考资料: https://leetcode.com/problems/score-of-parentheses/ https://leetcode.com/problems/score-of-parentheses/discuss/141777/C%2B%2BJavaPython-O(1)-Space