这道题在前两个的基础上做稍微改进就可以。
下面是AC代码：
1 /** 2 * Design an algorithm to find the maximum profit. You may complete at most two transactions. 3 * @param prices 4 * @return 5 */ 6 public int maxProfit3(int[] prices){ 7 if(prices==null || prices.length <= 1) 8 return 0; 9 int profit = 0; 10 int len = prices.length; 11 //the local minimum is marked as -1;the local maximum is marked as 1; otherwise is 0 12 int[] mark = new int[len]; 13 int[] left = new int[len];//mark whether the left is <= or >= 14 int[] right = new int[len];//mark whether the right is <= or >= 15 // the first number is initialized as 0 16 left[0] = 0; 17 //get the left array, by one traversal 18 int i=1; 19 while(i<len){ 20 if(prices[i-1]<prices[i]) 21 left[i] = -1; 22 else if(prices[i-1]>prices[i]) 23 left[i] = 1; 24 else 25 left[i] = left[i-1]; 26 i++; 27 } 28 //initialize the last element 29 right[len-1] = 0; 30 i = len-2; 31 while(i>=0){ 32 if(prices[i+1]<prices[i]) 33 right[i] = -1; 34 else if(prices[i+1]>prices[i]) 35 right[i] = 1; 36 else 37 right[i] = right[i+1]; 38 i--; 39 } 40 //get the mark 41 i = 0; 42 while(i<len){ 43 if(left[i] == right[i]){ 44 mark[i] = left[i]*(-1); 45 }else { 46 if(left[i] == 0){ 47 mark[i] = right[i]*(-1); 48 } 49 else if(right[i] == 0) 50 mark[i] = left[i]*(-1); 51 else 52 mark[i] = 0; 53 } 54 i++; 55 } 56 for( i=0;i<len-1;i++){ 57 if(mark[i] == 1){ 58 int temp1 = oneTransaction(0, i+1, prices); 59 int temp2=0; 60 if(i+1<len) 61 temp2 = oneTransaction(i+1,len,prices); 62 if(temp1+temp2>profit) 63 profit = temp1+temp2; 64 } 65 } 66 int temp = oneTransaction(0,len,prices); 67 if(temp>profit) 68 profit = temp; 69 return profit; 70 } 71 /** 72 * 73 * @param start [ 74 * @param end ） 75 * @param mark 76 * @param prices 77 * @return 78 */ 79 private int oneTransaction(int start, int end, int[] prices){ 80 if(end-start<=1) 81 return 0; 82 if(end-start==2) 83 return prices[end-1]-prices[start]>0?prices[end-1]-prices[start]:0; 84 int len = prices.length; 85 int[] lmin = new int[len]; 86 int[] rmax = new int[len]; 87 lmin[start] = prices[start]; 88 int i = start+1; 89 while(i<end){ 90 if(prices[i]>lmin[i-1]) 91 lmin[i] = lmin[i-1]; 92 else 93 lmin[i] = prices[i]; 94 i++; 95 } 96 i = end-1; 97 rmax[i] = prices[i]; 98 i--; 99 while(i>=start){ 100 if(prices[i]>rmax[i+1]) 101 rmax[i] = prices[i]; 102 else 103 rmax[i] = rmax[i+1]; 104 i--; 105 } 106 int max = 0; 107 for(i=start;i<end;i++){ 108 if(rmax[i]-lmin[i]>max) 109 max = rmax[i] - lmin[i]; 110 } 111 return max; 112 } 113