LeetCode 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 = 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.

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给定一个数字字符串，要求从中删除k个数字，使得最终结果最小，返回最小的数字字符串。 使用栈和贪心的思路。为了使结果尽量小，我们从左往右扫描字符串，把字符存到一个栈中，如果当前字符比栈顶小，则可以把栈顶数字删掉，这样相当于用当前字符代替栈顶字符所在位置，使得结果更小了。否则把当前字符压栈，注意此时不能把当前字符删掉，因为可能后面还有更大的字符出现。 如果这样过了一遍之后，还是没删够k个字符，此时，字符串从左往右肯定是非降序的。所以我们依次弹栈，把栈顶的元素删掉，直到删够k个字符。 最后还要判断一下剩余字符串是否为空，并删除前导零。完整代码如下： [cpp] class Solution { public: string removeKdigits(string num, int k) { if (k >= num.size())return "0"; string ans = ""; for (const auto& c : num) { if (ans.empty() || k == 0)ans.push_back(c); else { while (!ans.empty() && ans.back() > c) { ans.pop_back(); if (–k == 0)break; } ans.push_back(c); } } while (k– > 0 && !ans.empty()) ans.pop_back(); while (!ans.empty() && ans[0] == ‘0’)ans.erase(ans.begin()); return ans.empty() ? "0" : ans; } }; [/cpp] 本代码提交AC，用时6MS。]]>