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package class39;
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// 来自腾讯
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// 给定一个只由0和1组成的字符串S，假设下标从1开始，规定i位置的字符价值V[i]计算方式如下 :
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// 1) i == 1时，V[i] = 1
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// 2) i > 1时，如果S[i] != S[i-1]，V[i] = 1
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// 3) i > 1时，如果S[i] == S[i-1]，V[i] = V[i-1] + 1
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// 你可以随意删除S中的字符，返回整个S的最大价值
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// 字符串长度<=5000
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public class Code01_01AddValue {
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public static int max1(String s) {
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if (s == null || s.length() == 0) {
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return 0;
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}
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char[] str = s.toCharArray();
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int[] arr = new int[str.length];
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for (int i = 0; i < arr.length; i++) {
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arr[i] = str[i] == '0' ? 0 : 1;
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}
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return process1(arr, 0, 0, 0);
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}
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// 递归含义 :
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// 目前在arr[index...]上做选择, str[index...]的左边，最近的数字是lastNum
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// 并且lastNum所带的价值，已经拉高到baseValue
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// 返回在str[index...]上做选择，最终获得的最大价值
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// index -> 0 ~ 4999
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// lastNum -> 0 or 1
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// baseValue -> 1 ~ 5000
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// 5000 * 2 * 5000 -> 5 * 10^7(过!)
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public static int process1(int[] arr, int index, int lastNum, int baseValue) {
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if (index == arr.length) {
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return 0;
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}
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int curValue = lastNum == arr[index] ? (baseValue + 1) : 1;
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// 当前index位置的字符保留
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int next1 = process1(arr, index + 1, arr[index], curValue);
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// 当前index位置的字符不保留
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int next2 = process1(arr, index + 1, lastNum, baseValue);
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return Math.max(curValue + next1, next2);
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}
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// 请看体系学习班，动态规划章节，把上面的递归改成动态规划！看完必会
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}
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