暴力法超时:思想:动态规划 public int minFlipsMonoIncrb(String S) { int result = S.length(); for (int i = 0; i < S.length(); i++) { char[] str1 = S.substring(0, i).toCharArray(); char[] str2 = S.substring(i + 1, S.length()).toCharArray(); int zero = 0; int one =…
A string of '0's and '1's is monotone increasing if it consists of some number of '0's (possibly 0), followed by some number of '1's (also possibly 0.) We are given a string S of '0's and '1's, and we may flip any '0' to a '1' or a '1' to a '0'. Retu…
A string of '0's and '1's is monotone increasing if it consists of some number of '0's (possibly 0), followed by some number of '1's (also possibly 0.) We are given a string S of '0's and '1's, and we may flip any '0' to a '1' or a '1' to a '0'. Retu…
Given a non-negative integer N, find the largest number that is less than or equal to N with monotone increasing digits. (Recall that an integer has monotone increasing digits if and only if each pair of adjacent digits x and y satisfy x <= y.) Examp…
Given a non-negative integer N, find the largest number that is less than or equal to N with monotone increasing digits. (Recall that an integer has monotone increasing digits if and only if each pair of adjacent digits x and y satisfy x <= y.) Examp…
738. 单调递增的数字 知识点:字符串:贪心 题目描述 给定一个非负整数 N,找出小于或等于 N 的最大的整数,同时这个整数需要满足其各个位数上的数字是单调递增. (当且仅当每个相邻位数上的数字 x 和 y 满足 x <= y 时,我们称这个整数是单调递增的.) 示例 输入: N = 10 输出: 9 输入: N = 1234 输出: 1234 输入: N = 332 输出: 299 解法一:贪心 想一下这个过程,如果比前一位大的话那就可以不用动直接返回:如果比前一位小的话例如98,应该怎么做…
738. 单调递增的数字 给定一个非负整数 N,找出小于或等于 N 的最大的整数,同时这个整数需要满足其各个位数上的数字是单调递增. (当且仅当每个相邻位数上的数字 x 和 y 满足 x <= y 时,我们称这个整数是单调递增的.) 示例 1: 输入: N = 10 输出: 9 示例 2: 输入: N = 1234 输出: 1234 示例 3: 输入: N = 332 输出: 299 说明: N 是在 [0, 10^9] 范围内的一个整数. class Solution { public int…
