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| 1 | +package leetcode.str; |
| 2 | + |
| 3 | +/** |
| 4 | + * <p> |
| 5 | + * 给定一个整数,从整数中去掉k个数字,要求剩下的新整数尽可能小。给定的整数大小可以超过long类型的范围,所以需要用字符串来表示。 |
| 6 | + * </p> |
| 7 | + * |
| 8 | + * @author jwzhao |
| 9 | + * @version 1.0 |
| 10 | + * @date 2018/11/29 14:35 |
| 11 | + */ |
| 12 | +public class RemoveKDigits { |
| 13 | + |
| 14 | + public static void main(String[] args) { |
| 15 | + RemoveKDigits removeKDigits = new RemoveKDigits(); |
| 16 | + String num = "1432219"; |
| 17 | + int k = 3; |
| 18 | + System.out.println(removeKDigits.removeKDigits(num, k)); |
| 19 | + } |
| 20 | + |
| 21 | + public String removeKDigits(String num, int k) { |
| 22 | + if (num == null || num.length() == 0 || num.length() == k) { |
| 23 | + return "0"; |
| 24 | + } |
| 25 | + if (k == 0) { |
| 26 | + return num; |
| 27 | + } |
| 28 | + |
| 29 | + int newLength = num.length() - k; |
| 30 | + |
| 31 | + char[] nums = num.toCharArray(); |
| 32 | + Stack stack = new Stack(nums.length); |
| 33 | + for(int i=0;i<nums.length;i++){ |
| 34 | + String value = String.valueOf(nums[i]); |
| 35 | + String stackPop = stack.peek(); |
| 36 | + while (k>0 && stackPop != null && (Integer.parseInt(value) < Integer.parseInt(stackPop))) { |
| 37 | + // 栈顶弹出,入当前值 |
| 38 | + stack.pop(); |
| 39 | + stackPop = stack.peek(); |
| 40 | + k--; |
| 41 | + } |
| 42 | + stack.push(value); |
| 43 | + } |
| 44 | + |
| 45 | + StringBuffer buffer = new StringBuffer(); |
| 46 | + StringBuffer zero = new StringBuffer(); |
| 47 | + while (!stack.isEmpty()) { |
| 48 | + String value = stack.pop(); |
| 49 | + if ("0".equals(value)){ |
| 50 | + zero.append(value); |
| 51 | + }else{ |
| 52 | + if (zero.length() > 0){ |
| 53 | + buffer.append(zero); |
| 54 | + } |
| 55 | + buffer.append(value); |
| 56 | + zero = new StringBuffer(); |
| 57 | + } |
| 58 | + } |
| 59 | + if (buffer.length() == 0) { |
| 60 | + return "0"; |
| 61 | + } |
| 62 | + if (newLength > buffer.length()){ |
| 63 | + newLength = buffer.length(); |
| 64 | + } |
| 65 | + return buffer.reverse().substring(0,newLength); |
| 66 | + } |
| 67 | + |
| 68 | + public class Stack { |
| 69 | + |
| 70 | + /** |
| 71 | + * 栈底层实现的数组 |
| 72 | + */ |
| 73 | + private String[] stacks; |
| 74 | + /** |
| 75 | + * 栈顶指针 |
| 76 | + */ |
| 77 | + private int top; |
| 78 | + |
| 79 | + public Stack() { |
| 80 | + stacks = new String[8]; |
| 81 | + top = -1; |
| 82 | + } |
| 83 | + |
| 84 | + public Stack(int maxSize) { |
| 85 | + stacks = new String[maxSize]; |
| 86 | + top = -1; |
| 87 | + } |
| 88 | + |
| 89 | + /** |
| 90 | + * 入栈 |
| 91 | + * |
| 92 | + * @param val |
| 93 | + */ |
| 94 | + public void push(String val) { |
| 95 | + stacks[++top] = val; |
| 96 | + } |
| 97 | + |
| 98 | + /** |
| 99 | + * 查看栈顶元素 |
| 100 | + * |
| 101 | + * @return |
| 102 | + */ |
| 103 | + public String peek() { |
| 104 | + if (this.isEmpty()) { |
| 105 | + return null; |
| 106 | + } |
| 107 | + return stacks[top]; |
| 108 | + } |
| 109 | + |
| 110 | + /** |
| 111 | + * 弹出栈顶元素 |
| 112 | + * |
| 113 | + * @return |
| 114 | + */ |
| 115 | + public String pop() { |
| 116 | + if (this.isEmpty()) { |
| 117 | + return null; |
| 118 | + } |
| 119 | + return stacks[top--]; |
| 120 | + } |
| 121 | + |
| 122 | + /** |
| 123 | + * 判断是否为空 |
| 124 | + * |
| 125 | + * @return |
| 126 | + */ |
| 127 | + public boolean isEmpty() { |
| 128 | + return top == -1; |
| 129 | + } |
| 130 | + |
| 131 | + } |
| 132 | +} |
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