Algorithms—123.Best Time to Buy and Sell Stock III

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思路：限制交易两次。首先，求出单次交易最大利润，然后，只有2种可能，与此次交易无关，那么不在左侧就在右侧，再求一次，或者是与此次交易有关，那么在此次交易的日期限制内找出一段利润最低（负值）。比较左侧，右侧，中间的负值的绝对值的大小。

这段代码写的比较丑，凑合的看吧。sum3是中间的负值。

public class Solution {    public int maxProfit(int[] prices) {if (prices.length <= 1) {return 0;}int[] change = new int[prices.length - 1];for (int i = 0; i < change.length; i++) {change[i] = prices[i + 1] - prices[i];}int sum0 = 0;int s0 = 0;int b=0;int e=0;int begin0 = 0;int end0 = 0;for (int i = 0; i < change.length; i++) {s0 += change[i];if (s0 < 0) {s0 = 0;b=i+1;e=i+1;}if (change[i] > 0) {e=i;}if (s0 > sum0) {sum0 = s0;begin0=b;end0=e;}}int sum1 = 0;int s1 = 0;for (int i = 0; i <begin0; i++) {s1 += change[i];if (s1 < 0) {s1 = 0;}if (s1 > sum1) {sum1 = s1;}}int sum2 = 0;int s2 = 0;for (int i = end0+1; i <change.length; i++) {s2 += change[i];if (s2 < 0) {s2 = 0;}if (s2 > sum2) {sum2 = s2;}}int sum3 = 0;int s3 = 0;for (int i = begin0+1; i <end0; i++) {s3 += change[i];if (s3 > 0) {s3 = 0;}if (s3 < sum3) {sum3 = s3;}}return sum0+Math.max(0-sum3, Math.max(sum1, sum2));}}

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