Description:
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 = 3
Output: “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 = 1
Output: “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 = 2
Output: “0”
Explanation: Remove all the digits from the number and it is left with nothing which is 0.
解題思路及算法分析
本題是medium，比較麻煩的是要弄清楚其規律，很容易遺漏0的情況。考慮怎麼解決，就要先弄清楚去數規律：比如去k個數字，要是前k個裏面包含0，那麼就要把0留下，最後自動去除；若沒有0，要去的這個數，一定是從左到右，第一次比它後面的那個數字大的數字，但是若數字是非遞減的，則去掉後面最大的數字保留第一數的數即可。醬紫就可以解決所有情況，包括是0的情況。
注意：最後輸出結果的時候要去掉前面的0。
代碼
class Solution {
public:
string removeKdigits(string num, int k) {
int n = num.size();
if(k >= n) return "0";
while(k) {
int i = 0;
bool flag = false;
for (i = 0; i<num.size(); i++) {
if(num[i] > num[i+1]) {
flag = true;
num = num.erase(i,1);
break;
}
}
if(flag == false) {
num = num.substr(0,num.size()-1);
}
k--;
}
while(num.size() > 1 && num[0] == '0') {
num = num.substr(1);
}
return num;
}
};