402. Remove K Digits

Given a non-negative integer num represented as a string, remove k digits from the number so that the new number is the smallest possible.

Note:

• The length of num is less than 10002 and will be ≥ k.
• The given num does not contain any leading zero.

Example 1:

Input: num = "1432219", k = 3Output: "1219"Explanation: Remove the three digits 4, 3, and 2 to form the new number 1219 which is the smallest.

Example 2:

Input: num = "10200", k = 1Output: "200"Explanation: Remove the leading 1 and the number is 200. Note that the output must not contain leading zeroes.

Example 3:

Input: num = "10", k = 2Output: "0"Explanation: Remove all the digits from the number and it is left with nothing which is 0.

思路：怎样才能让结果最小呢？从前面开始，发现小的数字就应该把前面的数字替换掉，有个明显的出栈入栈的过程，于是乎用栈，那能确保怎样替换是正确的结果吗？仔细想想发现其实这就是greedy algorithm，是可以得到正确的结果的

/* * greedy algorithm using stack * 每一步都是到目前为止最优的结果(不管后面是什么)），谓之贪心 * 后一步的最优结果一定一前面的最优结果为基础 */public class Solution {    public String removeKdigits(String num, int k) {    if(num.length() == k)return "0";    Stack<Character> stack = new Stack<Character>();    char[] cs = num.toCharArray();    int cnt = k;        for(char c : cs) {    while(!stack.isEmpty() && stack.peek()>c && cnt>0) {    stack.pop();    cnt --;    }    stack.push(c);    }    StringBuilder sb = new StringBuilder();    while(!stack.isEmpty())sb.append(stack.pop());        //delete 0 from end    while(sb.length()>0 && sb.charAt(sb.length()-1) == '0')    sb.deleteCharAt(sb.length()-1);        int maxLen = Math.min(num.length()-k, sb.length());return sb.length()==0 ? "0" : sb.reverse().substring(0, maxLen).toString();    }}

二刷：

直观感觉贪心，策略是尽可能把小的数放到高位上，最佳的情况就是从左到右都是递增的，

但是如果不是呢？某一位比前面的数小，那我们可以肯定是要删除前面大的树，让这个更小的数往前面靠，

考虑到前面数是升序的，所以删除就是从后往前删，这不就是单调栈嘛