2021-04-25 09:10:55 +08:00
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//给定一个以字符串表示的非负整数 num,移除这个数中的 k 位数字,使得剩下的数字最小。
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//
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// 注意:
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//
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//
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// num 的长度小于 10002 且 ≥ k。
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// num 不会包含任何前导零。
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//
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//
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// 示例 1 :
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//
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//
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//输入: num = "1432219", k = 3
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//输出: "1219"
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//解释: 移除掉三个数字 4, 3, 和 2 形成一个新的最小的数字 1219。
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//
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//
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// 示例 2 :
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//
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//
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//输入: num = "10200", k = 1
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//输出: "200"
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//解释: 移掉首位的 1 剩下的数字为 200. 注意输出不能有任何前导零。
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//
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//
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// 示例 3 :
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//
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//
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//输入: num = "10", k = 2
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//输出: "0"
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//解释: 从原数字移除所有的数字,剩余为空就是0。
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//
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// Related Topics 栈 贪心算法
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// 👍 556 👎 0
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package leetcode.editor.cn;
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import java.util.Stack;
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//402:移掉K位数字
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public class RemoveKDigits {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new RemoveKDigits().new Solution();
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//1219
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System.out.println(solution.removeKdigits("1432219", 3));
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//200
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System.out.println(solution.removeKdigits("10200", 1));
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//0
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System.out.println(solution.removeKdigits("10", 2));
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//0
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System.out.println(solution.removeKdigits("10", 1));
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//11
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System.out.println(solution.removeKdigits("112", 1));
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public String removeKdigits(String num, int k) {
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if (num.length() == k) {
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return "0";
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}
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Stack<Character> stack = new Stack<>();
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int remove = 0;
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int i = 0;
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for (; i < num.length() - 1; i++) {
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while (i + 1 < num.length() && num.charAt(i) > num.charAt(i + 1) && remove < k) {
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num = num.substring(i + 1);
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2021-04-25 09:42:28 +08:00
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i=0;
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2021-04-25 09:10:55 +08:00
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remove++;
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2021-04-25 09:42:28 +08:00
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while (!stack.isEmpty() && i < num.length() && stack.peek() > num.charAt(i) && remove < k) {
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2021-04-25 09:10:55 +08:00
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stack.pop();
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remove++;
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}
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}
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if (remove == k) {
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break;
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}
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if (i < num.length()) {
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stack.push(num.charAt(i));
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}
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}
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2021-04-25 09:42:28 +08:00
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num = num.substring(i);
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2021-04-25 09:10:55 +08:00
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while (!stack.isEmpty()) {
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num = stack.pop() + num;
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}
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if (remove < k) {
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num = num.substring(0, num.length() - k + remove);
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}
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while (num.length() > 1 && num.startsWith("0")) {
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num = num.substring(1);
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}
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return num;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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